Topics

Statement : \(\ dQ = dU + dW\ \) (For zero heat loss)
\(dQ =\) heat supplied to the system
\(dU =\) change in internal energy
\(dW =\) work done by the system

• This law is a form of energy conservation.

The first Law of Thermodynamics (derivation)
Let us introduce certain amount of heat energy dQ into a gas confined inside a cylinder fit with a piston.
The gas can either:
(i) store energy as random KE of its molecules \((dU)\), or
(ii) use the energy to do work ( dW ) in the environment (such as raising a weight on the piston).

Thus, the first law of thermodynamics can be written : \(dQ = dU + dW\)
Note: In solving thermodynamics problems, always take gas as system.

Sign convention :

(i) Whenever heat is added to the system sign is +ve
(ii) Whenever system rejects the heat sign is -ve

Work (dW) :

(i) \((V \uparrow )\) sign is +ve
(ii) sign is -ve (V \(\downarrow\) )

Internal energy (dU) :

(i) When temperature increases sign is +ve
(ii) When temperature decreases sign is -ve

Isochoric process :

A process that takes place at constant volume is called isochoric process or isovolumetric process. (Gas will follow Gaylusac's Law)
(i) Conditions \(\longrightarrow V\) is constant
(ii) Process equation \(\longrightarrow \frac{P}{T} =\) constant
(iii) If gas is taken from state A having pressure \(P_{1}\) temperaure \(T_{1}\) and volume V to state B having Pressure \(P_{2}\) temperaure \(T_{2}\) and volume \(V\) then
\[{\underset{\left( P_{1},\text{ }V,{\text{ }T}_{1} \right)}{A} \longrightarrow \underset{\left( P_{2},\text{ }V,{\text{ }T}_{2} \right)}{B} }{\frac{P_{1}}{{\text{ }T}_{1}} = \frac{P_{2}}{{\text{ }T}_{2}} }\][While \(PV = nRT\) will relate variables in same state ]
(iv) \(P - V\) curve

(v) \(dW = 0\) since \(dV = 0\)
(vi) \(\ \Delta U = \frac{nfR}{2}\left( {\text{ }T}_{2} - T_{1} \right)\)
(vii) \(\ \Delta Q = {nC}_{v}\left( T_{2} - T_{1} \right)\)
(viii) Applying first law of thermodynamics (F.L.T.)

\[\begin{matrix} & O + \Delta U = \Delta Q \\ \Rightarrow \ & n\left( \frac{fR}{2} \right)\left( T_{2} - T_{1} \right) = {nC}_{v}\left( {\text{ }T}_{2} - T_{1} \right) \\ \Rightarrow \ & C_{v} = \frac{fR}{2} \end{matrix}\]

(ix) Bulk modulus \(= - \frac{dP}{(dV/V)} = - \infty\)
(x) sp. heat \(= C_{v}\)

Isobaric process :

A process that takes place at constant pressure is called isobaric process. (Gas will follow Charle's Law)
(i) Condition : P is constant
(ii) Process equation \(\longrightarrow \frac{V}{T}\) is constant
(iii) If gas is taken from state A having pressure P temperaure \(T_{1}\) and volume \(V_{1}\) to state B having Pressure P temperaure \(T_{2}\) and volume \(V_{2}\) then
\[{\underset{\phantom{\left( P,V_{1},{\text{ }T}_{1} \right)\ }}{\overset{\phantom{A\ }}{\rightarrow}}\ \underset{\left( P,V_{2},{\text{ }T}_{2} \right)}{B} }{\frac{V_{1}}{{\text{ }T}_{1}} = \frac{V_{2}}{{\text{ }T}_{2}}}\]

\[\begin{matrix} & {PV}_{1} = {nRT}_{1} \\ & {PV}_{2} = {nRT}_{2} \end{matrix}\]

(iv)

(v) \(\ \Delta W = \int_{v_{1}}^{v_{2}}\mspace{2mu} Pdv\)
\[{= P\left( V_{2} - V_{1} \right) }{= nR\left( T_{2} - T_{1} \right) }\](vi) \(\ dU = {nC}_{v}dT\)
\[\Delta U = {nC}_{v}\left( T_{2} - T_{1} \right) \](vii) \(dQ = {nC}_{p}dT\)
\[\Delta Q = {nC}_{p}\left( T_{2} - T_{1} \right) \](viii) Applying FLT

\[\begin{matrix} & {nC}_{p}\left( {\text{ }T}_{2} - T_{1} \right) = {nC}_{v}\left( {\text{ }T}_{2} - T_{1} \right) + nR\left( {\text{ }T}_{2} - T_{1} \right) \\ \Rightarrow \ & C_{p} - C_{v} = R \end{matrix}\]

(known as meyer's relation)

Gas constant expressed for particular gas in terms of mass is known as characteristic gas constant of that particular gas.

\[R_{\text{characteristic}\text{~}} = \frac{R\text{~}\text{(universal gas const.)}\text{~}}{M\text{~}\text{(}\text{mol.wt.of}\text{ gas)}\text{~}}\]

Units J/gm-k or J/kg-k
(ix) Bulk modulus

\[B = \frac{- dp}{(dV/V)} = 0\]

(x) Sp. heat capacity \(C_{p}\)

Isothermal process :

A process that takes place at constant temperature is called isothermal process. (Gas will follow Boyle's Law)
(i) Condition \(\longrightarrow\) Temp. is constant
(ii) Process equation \(\longrightarrow PV =\) const.
(iii) If gas is taken from state \(A\) having pressure \(P_{1}\) volume \(V_{1}\) and temperaure \(T\) to state \(B\) having Pressure \(P_{2}\) volume \(V_{2}\) and temperaure \(T\) then

\[\begin{matrix} P_{1}V_{1} = P_{2}V_{2} & P_{1}V_{1} = nRT \\ & P_{2}V_{2} = nRT \end{matrix}\]

(iv) P-V curve

\[\text{~}\text{slope}\text{~} = \frac{dP}{dV} = \frac{- P}{\text{ }V}\]

(v) \(\ \Delta W = \int_{V_{1}}^{V_{2}}\mspace{2mu} PdV = k\int_{V_{1}}^{V_{2}}\mspace{2mu}\frac{dV}{V}\)

\[\begin{matrix} & \ = PVln\frac{{\text{ }V}_{2}}{{\text{ }V}_{1}} \\ & \ = P_{1}{\text{ }V}_{1}ln\frac{{\text{ }V}_{2}}{{\text{ }V}_{1}} = P_{1}{\text{ }V}_{2}ln\frac{{\text{ }V}_{2}}{{\text{ }V}_{1}} \\ & \ = nRTln\frac{{\text{ }V}_{2}}{{\text{ }V}_{1}} = nRTln\frac{P_{1}}{P_{2}} \end{matrix}\]

(vi) \(\ dU = 0\)
(vii) Using FLT

\[\begin{matrix} & dQ = dW \\ & \Delta Q = nRTln\frac{{\text{ }V}_{2}}{{\text{ }V}_{1}} \end{matrix}\]

(viii) \(B = - \frac{dP}{(dV/V)} = P\)
(ix) Specific heat \(= C = \left( \frac{dQ}{ndT} \right) = \infty\)

By adiabatic process :

Adiabatic Process (Derivation): For an adiabatic process ( \(\Delta Q = 0\) )

\[\begin{matrix} & 0 = C_{v} + \frac{P}{n}\frac{dV}{dT} = \frac{R}{\gamma - 1} + \frac{RT}{\text{ }V}\frac{dV}{dT}\ \Rightarrow \ (\gamma - 1)\int_{}^{}\ \frac{dV}{\text{ }V} + \int_{}^{}\ \frac{dT}{\text{ }T} = 0 \\ & \ (\gamma - 1)lnV + lnT = lnc^{'}\therefore{TV}^{\gamma - 1} = c^{'};\text{~}\text{Also,}\text{~}T = \frac{PV}{nR}\therefore{PV}^{'} = c^{'} \\ & P\left( \frac{\text{ }T}{P} \right)^{\gamma} = c^{'}\therefore\frac{T^{\gamma}}{P^{\gamma - 1}} = c^{'}\ \therefore P = T^{(\gamma\gamma - 1)} \end{matrix}\]

In this process net heat supplied to the gas is zero
(i) Condition \(dQ = 0\)
(ii) Process equation \({PV}^{r} =\) const.
(iii) If gas is taken from state \(A\) having pressure \(P_{1}\) volume \(V_{1}\) and temperaure \(T_{1}\) to state \(B\) having Pressure \(P_{2}\) volume \(V_{2}\) and temperaure \(T_{2}\) then

\[P_{1}{\text{ }V}_{1}^{\gamma} = P_{2}{\text{ }V}_{2}^{\gamma}\ P_{1}{\text{ }V}_{1} = {nRT}_{1}\ P_{2}{\text{ }V}_{2} = {nRT}_{2}\]

(iv) \(P - V\) curve

\[\text{~}\text{slope}\text{~} = \frac{dP}{dV} = \frac{- \gamma P}{V}\]

(v) \(\ \Delta W = \int_{V_{1}}^{V_{2}}\mspace{2mu} PdV\left( \right.\ \) since \(\left. \ {Pv}^{t} = k \right)\)

\[= k\int_{v_{1}}^{v_{2}}\mspace{2mu}{\text{ }V}^{- \gamma}dV\ = \left. \ \frac{{kv}^{- \gamma + 1}}{( - \gamma + 1)} \right|_{v_{1}}^{v_{2}} = \frac{P_{2}{\text{ }V}_{2} - P_{1}{\text{ }V}_{1}}{1 - \gamma}\]

(vi) \(\ \Delta V = {nC}_{v}\left( T_{2} - T_{1} \right)\)

\[= \frac{nR\left( {\text{ }T}_{2} - T_{1} \right)}{\gamma - 1} = \frac{P_{2}{\text{ }V}_{2} - P_{1}{\text{ }V}_{1}}{\gamma - 1}\]

(vii) Applying FLT we can see that

\[0 = \Delta W + \Delta U \Rightarrow \Delta U = - \Delta W\]

(viii) \(\ B = - \frac{dP}{(dV/V)} = \gamma P\)
(ix) Sp. heat

\[C = \frac{dQ}{ndT} = 0\]

Results :

(i) For expansion from same initial vol. & pressure to same final volume \(\rightarrow\)

As the area under the \(P - V\) graph is largest for isobaric hence its work done is greater than isothermal and in similar manner we can say that work done in isothermal process will be greater than adiabatic process.
(ii) For compression from same initial state to same final volume

(iii) At pt. of interaction (slope of isothermal) \(\times \gamma =\) slope of adiabatic.

It is a process in which molar heat capacity \(C =\) constant

\[C = C_{v} + \frac{P}{n}\frac{dV}{dT};\ \ PdV + VdP = nRdT\]

\[{\Rightarrow \ C = C_{v} + \frac{PRdV}{PdV + VdP} }{\Rightarrow \ \frac{R}{C - C_{v}} = 1 + \frac{VdP}{PdV} }{\Rightarrow \ \frac{R - \left( C - C_{v} \right)}{C - C_{v}} = \frac{VdP}{PdV} = \frac{C - C_{p}}{C - C_{v}} = K }{\Rightarrow \ \int\frac{dP}{P} + K\int\frac{dV}{V} = 0 \Rightarrow \ lnP + KlnV =}\] const
\(\therefore\ {PV}^{K} =\) const. \(\ K\) : Polytropic constant

Isochoric: \(\ K = \infty\ \left\lbrack P^{1/K}V = \right.\ \) const. \(\rbrack\)
\[C = C_{v} \]Isocbaric: \(\ K = 0\)
\[C = C_{p} \]Isothermal: \(K = 1\ \text{ }K = \frac{C - C_{p}}{C - C_{v}} = \frac{1 - C_{p}/C}{1 - C_{v}/C}\ C = \infty\)
Adiabatic: \(K = \gamma\ \lbrack\) Put \(C = o\) in the equation of K\(\rbrack\ C = 0\)
(i) Characteristic equation process

\[{PV}^{x} = \text{~}\text{const}\text{~}.\]

(ii) \(\Delta W = \int_{V_{1}}^{V_{2}}\mspace{2mu} PdV\)

\[\begin{matrix} & \ = \int_{V_{1}}^{V_{2}}\mspace{2mu}\mspace{2mu}\frac{k}{V^{x}}dV \\ & \ = \frac{\frac{k}{V^{x}}V_{2} - \frac{k}{V_{1}^{x}}V_{1}}{1 - x}1 = \frac{P_{2}V_{2} - P_{1}V_{1}}{1 - x} \\ & \ = \frac{nR\left( T_{2} - T_{1} \right)}{1 - x + 1} - V_{1}^{- x + 1} \\ & \end{matrix}\]

(iii) \(\Delta U = \frac{nR\left( T_{2} - T_{1} \right)}{(\gamma - 1)}\)
(iv) From FLT

\[\begin{matrix} & dQ = dW + dU \\ & \Delta Q = nR\left( {\text{ }T}_{2} - T_{1} \right)\left\lbrack \frac{1}{1 - x} + \frac{1}{\gamma - 1} \right\rbrack \end{matrix}\]

(v) Sp. heat

\[\begin{matrix} & \Delta Q = nc\left( {\text{ }T}_{2} - T_{1} \right) \\ & C = \frac{R}{1 - x} + \frac{R}{\gamma - 1} \end{matrix}\]

It is combination of two or more than two processes in which initial and final state of the system is same
(i) \(\ dU = 0\)

Total change in I.E. is zero because initial & final temp. is same
(ii) Applying FLT

\[dQ = dW\]

(iii) Net work done in a cyclic process is area bounded by process curve on PV diagram.

(iv) Efficiency of cyclic process :

\[\eta = \frac{\text{~}\text{Total work done (}\text{~} + \text{~}\text{ve}\text{ and}\text{~} - \text{~}\text{ve}\text{ both)}\text{~}}{\text{~}\text{Total heat supplied (only}\text{~} + \text{~}\text{ve}\text{ heat) to the system}\text{~}} \times 100\]

(A) The Engine :

As according to the second law of thermodynamics whole of heat can never be converted into work, the question then arises under what conditions the conversion of heat to work is these questions Carnot developed an ideal heat engine which is supposed to consist of the following four components :
(1) A cylinder with perfectly non-conducting walls and a perfectly conducting base containing a perfect gas as working substance and fitted with a non-conducting frictionaless piston'
(2) A source of infinite thermal capacity maintained at constant higher temperature \(T_{H}\);
(3) A sink of infinite thermal capacity maintained at constant lower temperature \(T_{L}\); and
(4) A perfectly non-conducting stand for the cylinder.

Here it is worth mentioning that as all the above mentioned components cannot exist in reality, Carnot engine is an ideal (hypothetical) engine which can never be actually constructed.
(B) Carnot Cycle [or Working of the Engine] :

The working substance in a Carnot engine is taken through a reversible cycle consisting of the folloing four steps :
(i) The cylinder containing ideal gas is placed on the source and the gas is allowed to expand slowly at constant temperature \(T_{H}\) absorbing heat

\(Q_{H}\). This isothermal change is represented by the curve \(AB\) in the indicator diagram.
(ii) The cylinder is then placed on the non-conducting stand and the gas is alloweed to expand adiabatically till the temperature falls from \(T_{H}\) to \(T_{L}\). This adiabatic expansion is represented by the curve \(BC\).
(iii) The cylinder is next placed on the sink and the gas is compressed at constant temperature \(T_{L}\). This adiabatic expansion is represented by the curve BC .
(iv) Finally the cylinder is again placed on the non-conducting stand and the compression is continued so that the gas returns to its initial stage along DA .
The closed pathABCDA represents the so called Carnot cycle and the four stages taken together represent a cyclic process
(C) Efficiency of the Engine :

The efficiency of a engine is defined as the ratio of work done to the heat supplied, i.e.,

\[\begin{array}{r} \eta = \frac{\text{~}\text{Work done}\text{~}}{\text{~}\text{heat input}\text{~}} = \frac{W}{W_{H}}\#(i) \end{array}\]

But as for cyclic process \(\Delta W =\) i.e., \(W = Q_{H} - Q_{L}\)
So eqn. (i) reduces to

\[\begin{array}{r} \eta = \frac{Q_{H} - Q_{L}}{Q_{H}} = 1 - \frac{Q_{L}}{Q_{H}}\#(ii) \end{array}\]

Now as in an isothermal process internal energy remains constant, in accordance with first law

\[\Delta Q = \Delta W = nRT\log_{e}\left( {\text{ }V}_{F}/V_{l} \right)\ \lbrack\text{~}\text{as}\text{~}\Delta U = 0\rbrack\]

So \(\ Q_{H} = {nRT}_{H}\log_{e}\left( \frac{V_{B}}{V_{A}} \right)\)
and

\[\begin{array}{r} \left| Q_{L} \right| = {nRT}_{L}log\left( \frac{{\text{ }V}_{C}}{R_{D}} \right)\#(iii) \end{array}\]

But as for adiabatics BC and DA respectively,

\[T_{H}{\text{ }V}_{B}^{\gamma - 1} = T_{L}{\text{ }V}_{C}^{\gamma - 1}\text{~}\text{and}\text{~}T_{H}{\text{ }V}_{A}^{\gamma - 1} = T_{L}{\text{ }V}_{D}^{\gamma - 1}\]

Dividing these two results,

\[\begin{array}{r} \left( \frac{V_{B}}{V_{A}} \right)^{\gamma - 1} = \left( \frac{V_{C}}{V_{D}} \right)^{\gamma - 1}\text{, i.e.,}\text{~}\frac{V_{B}}{V_{A}} = \frac{V_{C}}{V_{D}}\#(iv) \end{array}\]

Substituting the value of \(Q_{H}\) and \(Q_{L}\) from, equation (iii) in (ii) in the light of (iv), we get

\[\begin{array}{r} \eta = 1 - \frac{T_{L}}{{\text{ }T}_{H}}\#(v) \end{array}\]

This is the required result and from this it is clear that :
(1) Efficiency of a heat engine depends only on temperatures of source and sink and is independent of all other factors.
(2) All reversible heat engines working between same temperatures are equally efficient and no heat engine can be more efficient than Carnot engine (as it is ideal).
(3) As on Kelvin scale temperature can never be negative (as 0 K is defined as lowest possible temperature) and \(T_{H}\) and \(T_{L}\) are finite, efficiency of a heat engine is always lesser than unity, i.e., whole of heat can never be converted into work which is in accordance with second law.
The efficiency of actual engines is much lesser than that of ideal engine. Actually the practical efficiency of a steam engine is about \((8 - 15)\%\) while that of a petrol engine \(40\%\). The efficiency of a diessel engine is maximum and is about \((50 - 55)\%\).

Practice Exercise

Q. 1 The average energy and the rms speed of molecules in a sample of oxygen gas at 300 K are \(6.21 \times 10^{-}\ ^{21}\text{ }J\) and \(484{\text{ }ms}^{- 1}\) respectively. The corresponding values of 600 K are nearly
(A) \(12.42 \times 10^{- 21}\text{ }J,968{\text{ }ms}^{- 1}\)
(B) \(8.78 \times 10^{- 21}\text{ }J,684{\text{ }ms}^{- 1}\)
(C) \(6.21 \times 10^{- 21}\text{ }J,968{\text{ }ms}^{- 1}\)
(D) \(12.42 \times 10^{- 21}\text{ }J,684{\text{ }ms}^{- 1}\)
Q. 2 The P - T diagram for a ideal gas is shown in Figure where AC is an adiabatic process. The corresponding PV diagram is .

(A)

(B)

(C)

(D)

Q. 3 Variation of molar specific heat of a metal with temperature is best depicted by

(A)

(B)

(C)

(D)

Answers

Q. 1 D
Q. 2 B
Q. 3 D

Q. 170 calorie of heat is required to raise the temperture of a diatomic gas at contant pressure from 30 to \(35^{\circ}C\). the amount of heat requried (in calorie) to raise the temperature of the same gas through the same range ( 30 to \(35^{\circ}C\) ) at constant volume is
(A) 30
(B) 60
(C) 50
(D) 70

Sol. \(\frac{C_{P}}{C_{V}} = \gamma = 1.4\)
\[{\frac{(\Delta Q)_{P}}{(\Delta Q)_{V}} = \frac{{nC}_{p}\Delta T}{{nC}_{V}\Delta T} = \frac{C_{P}}{C_{V}} = 1.4 }{\therefore\ (\Delta Q)_{V} = \frac{(\Delta Q)_{P}}{1.4} = 50cal }\]Q. 2 A vessel contains 1 mole of \(O_{2}\) and 1 mole of He . The value of \(\gamma\) of the mixture is
(A) 1.4
(B) 1.50
(C) 1.53
(D) none of these

Sol. \(\ C_{\text{vix}\text{~}} = \frac{\frac{3}{4}R + \frac{5}{2}R}{2} = 2R\)
\[{C_{\text{pmix}\text{~}} = 2R + R = 3R }{y_{\text{mix}\text{~}} = \frac{C_{p}}{C_{V}} = \frac{3}{2} }\]Q. 3 An ideal is taken through a cycle \(A \rightarrow B \rightarrow C \rightarrow A\) as shown in Figure if the net heat supplied in the cycle is 5 J , then work done by the gas in the process \(C \rightarrow A\) is

(A) -5 J
(B) -10 J
(C) -15 J
(D) -20 J

Sol. Work done \(=\) area under the curve \(10\text{ }J;5 = W_{CA} + 10\) or \(W_{CA} = - 5\text{ }J\)
Q. 4 A glass tube scaled at both ends is 1 m long. It lies horizontally with the middle 10 cm containing Hg . The two ends of the tube, eqaul in length, contain air at \(27^{\circ}C\) and pressure 76 cm of Hg . The temperature at one end is kept \(0^{\circ}C\) and at the other end it \(127^{\circ}C\). Neglect the change in length of Hg column. Then the change in length on two sides is
(A) 12.3 cm
(B) 10.311 cm
(B) 9.9 cm
(D) 8.49 cm

Sol. Initially \(l = 45\text{ }cm(2l + 10 = 100\text{ }cm)\)

\[\begin{array}{r} P_{1} = P_{2} = P(\text{~}\text{say}\text{~})\#(i) \end{array}\]

(a)

(b)

Applying gas law at end A,

\[\begin{array}{r} \frac{45AP}{300} = \frac{(45 - x){AP}_{1}}{273}\#(ii) \end{array}\]

At end \(B\frac{45AP}{300} = \frac{(45 + x){AP}_{2}}{400}\)

From (i), (ii) and (iii)
\[\frac{(45 - x)}{273} = \frac{45 + x}{400} = 8.49\text{ }cm \]Q. 5 Find the amont of work done to increase the temperature of one mole of an ideal gas by \(30^{\circ}C\) if it is expanding under the condition \(V \propto T^{2/3}\).
(A) 166.2 J
(B) 136.2 J
(C) 126.2 J
(D) None of these

Sol. \(\ PV = RT\) for 1 mole

\[\begin{matrix} & W = \int_{}^{}\ PdV = \int_{}^{}\ \frac{RT}{\text{ }V}dV \\ & \text{ }V = {CT}^{2/3} \\ \therefore & dV = \frac{2}{3}{CT}^{2/3}dT \\ \text{~}\text{or}\text{~} & \frac{dV}{\text{ }V} = \frac{2}{3}\frac{dT}{\text{ }T} \\ \therefore & \text{ }W = \int_{T_{1}}^{T_{2}}\mspace{2mu}\mspace{2mu} RT\left( \frac{2}{3} \right)\frac{dT}{\text{ }T} \\ & = \frac{2}{3}R\left( {\text{ }T}_{2} - T_{1} \right) = 166.2\text{ }J \end{matrix}\]

Q. 6 A mercury thermometer read \(80^{\circ}C\) when the mercury is at 5.2 cm mark and \(60^{\circ}C\) when the mercury is at 3.9 cm mark. Find the temperature when the mercury level is at 2.6 cm mark.

Sol. \(\ \frac{l_{1} - l_{2}}{l_{1} - l_{3}} = \frac{\alpha l_{0}\left( {\text{ }T}_{1} - T_{2} \right)}{\alpha l_{0}\left( {\text{ }T}_{1} - T_{3} \right)} = \frac{T_{1} - T_{2}}{{\text{ }T}_{1} - T_{3}}\)
or \(\ \frac{5.2 - 3.9}{5.2 - 2.6} = \frac{80 - 60}{80 - T_{3}}\)
\(1.3\left( 80 - T_{3} \right) = 2.6(20)\) or \(T_{3} = 40^{\circ}C\)
Q. 7 A vertical cylinder pistion system has cross-section S . It contains 1 mole of an ideal monoatomic gas under a pistion of mass M . At a certain instant a heater is switched on which transmits a heat q per unit time ot the cylinder. Find the velocity \(v\) of the pistion under the condition that pressure under the pistionis constant and the system is thermally insulated.

Sol. Gas pressure \(= P_{0} + \frac{Mg}{S}\), Where M is mass of the pistion.
As C \(C_{v} = \frac{3}{2}R\)
\[{\therefore\Delta U = \frac{3}{2}R\Delta T = \frac{3}{2}P\Delta V }{Q = P\Delta V + \Delta U = P\Delta V + \frac{3}{2}P\Delta V = \frac{5}{2}P\Delta V }{\Delta V = Sdx }\]or \(Q = q \cdot dt = \frac{5}{2}PSdx\)
or \(\frac{dx}{dt} = \frac{2q}{5PS} = \frac{2q}{5\left( P_{0} + \frac{Mg}{S} \right)S}\)
Q. 8 A typer pumped to a pressure 3.3375 atm at \(27^{\circ}C\) suddenly bursts. What is the final temperature ( \(\gamma = 1.5)\) ?
(A) \(27^{\circ}C\)
(B) \(- 27^{\circ}C\)
(C) \(0^{\circ}C\)
(D*) \(- 73^{\circ}\)

Sol. \(\ T_{1}^{\gamma}P_{1}^{1 - \gamma} = T_{2}^{\gamma}P_{2}^{1 - \gamma}\)
or \(\left( \frac{T_{1}}{{\text{ }T}_{2}} \right)^{\gamma} = \left( \frac{P_{1}}{P_{2}} \right)^{\gamma - 1} = \left( \frac{300}{{\text{ }T}_{2}} \right)^{3/2} = \left( \frac{3.375}{1} \right)^{3/2 - 1}\)
or \(T_{2} = \frac{300}{(3.375)^{1/3}} = 200\text{ }K = - 73^{\circ}C\)
Q. 93 Moles of a gas mixture having volume V and temperature T is compressed to \(1/5\) th of the initial volume. Find the change in its adiabatic compressibility if the gas obeys \({PV}^{19/13} =\) constant \(\lbrack R = 8.3\text{ }J/mol - K\) ]
Sol. Bulk modulus B \(= \gamma P\)

\[\text{~}\text{Compressibility}\text{~}C = \left( \frac{1}{\text{ }B} \right) = \frac{1}{\gamma P}\]

and \(\ \Delta C = C - C\)
or \(\ DC = \frac{1}{\gamma}\left\lbrack \frac{1}{P^{'}} - \frac{1}{P} \right\rbrack\)

\[{PV}^{\gamma} = P^{'}\left( \frac{V}{5} \right)^{\gamma}\]

With \(\ \gamma = \frac{19}{13}\) and \(P^{'} = 5^{\gamma}P,11\)

\[\Delta C = \frac{1}{\gamma P}\left\lbrack \frac{1}{5^{y}} - \frac{1}{1} \right\rbrack = \frac{13 \times 0.905}{19P}\]

But \(\ PV = nRT\) or \(P = \frac{nRT}{V}\)

\[\Delta C = \frac{13.(0905)V}{19 \times 3 \times 8.317\text{ }T} = \frac{- 0.0248\text{ }V}{\text{ }T}\]

Q. 10 One mole of an ideal gas is contained under a weightless pistion of a vertical cylinder at a temperature T. The space over the pistion opens into the atmosphere. What work has to be performed in order to increaseisothermally the gas volume under the pistion \(\eta\) times
Sol. Let A be the area of cross-section

\[\begin{matrix} & F + PA = P_{0}\text{ }A \\ & \text{ }F = \left( P_{0} - P \right)A \end{matrix}\]

Work done by the agent
\[W = \int_{V}^{\eta V}\mspace{2mu} Fdx = \int_{V}^{\eta V}\mspace{2mu}\left( P_{0} - P \right)Adx \]

\[{= \int_{V}^{\eta V}\mspace{2mu}\left( P_{0} - P \right)dV }{= P_{0}(\eta - 1)V - \int_{V}^{\eta V}\mspace{2mu} nRT\frac{dV}{V} }{= RT(\eta - 1) - {nlog}_{e}\eta }\]Q. 11 Calculate the heat absorbed by the system in going through the process shown in figure
(A) 31.4 J
(B) 3.14 J
(C) \(3.14 \times 10^{4}\text{ }J\)
(D) none

Sol. \(\ \) Heat absorbed \(= \pi r^{2}\)
\[{= \pi\left( P_{r} \right)\left( V_{r} \right) }{= 3.14\left( 100 \times 10^{3} \right)\left( 100 \times 10^{- 6} \right) }{= 31.4\text{ }J }\]Q. 12 A mercury manometer consists of two unequal arms of equal cross-section \(1{\text{ }cm}^{2}\) and lengths 100 cm and 50 cm . The two open ends are sealed with air in the tube at a pressure of 80 cm of mercury. Some amount of mercury is now introduced in the monometer through the stopcock connected to it. If mercury resis in the shorter tube to a length 10 cm in steady state, find the length of the mercury column risen in the longer tube.
Sol. Let \(p_{1}\) and \(p_{2}\) and be the pressures in centimetre of mercury in the two arms after introducing mercury in the tube. Suppose the mercury column rises in the second arm to \(l_{0}\text{ }cm\).

Using \(pV =\) constant for the shorter arm ,

\[\begin{array}{r} (80\text{ }cm)(50\text{ }cm) = p_{1}(50\text{ }cm - 10\text{ }cm)\#(i) \end{array}\]

or \(\ p_{1} = 100\text{ }cm\).

Using \(pV =\) constant for the longer arm,

\[\begin{array}{r} (80\text{ }cm)(100\text{ }cm) = p_{2}\left( 100 - l_{0} \right)cm\#(ii) \end{array}\]

From the figure,

\[p_{1} = p_{2} + \left( l_{0} - 10 \right)cm\]

Thus by (i),

\[100\text{ }cm = p_{2} + \left( l_{0} - 10 \right)cm.\]

or \(\ p_{2} = 110\text{ }cm - l_{0}\text{ }cm\)
Putting in (ii),

\[\left( 110 - l_{0} \right)\left( 100 - l_{0} \right) = 8000\]

or, \(\ l_{0}\ ^{2} - 210l_{0} + 3000 = 0\)
or \(\ l_{0} = 15.5\)
The required length is 15.5 cm .
Q. 13 An ideal gas has pressure \(p_{0}\), volume \(V_{0}\) and temperature \(T_{0}\). It is taken through an isochoric process till its pressure is doubled. It is now isothermally expanded to get the original pressure. Finally, the gas is isobarically compressed to its original volume \(V_{0^{*}}\). (a) Show the process on a p-V diagram. (b)What is the temperature in the isothermal part of the process? (c) What is the volume at the end of the isothermal part of the process?
Sol. (a) The process is shown in a p-V diagram in figure. The process starts from A and goes through ABCA.

(b) Applying \(\ pV = nRT\) at A and B ,

\[p_{0}{\text{ }V}_{0} = {nRT}_{0}\]

and \(\ \left( 2p_{0} \right)V_{0} = {nRT}_{B}\)
Thus, \(\ T_{B} = 2{\text{ }T}_{0}\)
This is the temperature in the isothermal part BC .
(c) As the process BC is isothermal, \(T_{C} = T_{B} = 2{\text{ }T}_{0}\).

Applying \(pV = nRT\) at A and C ,

\[p_{0}{\text{ }V}_{0} = {nRT}_{0}\]

and

\[\begin{matrix} & p_{0}{\text{ }V}_{c}nR\left( 2{\text{ }T}_{0} \right) \\ & V_{c}2{\text{ }V}_{c^{*}} \end{matrix}\]

Q. 14 Figure shown a vertical cylindrical vessel separated in two parts by a frictionless pistion free to move along the length of the vessel. The length of the cylinder is 90 cm and the pistion divides the cylinder in the ratio of \(5:4\). Each of the two parts of the vessel contains 0.1 mole of an ideal gas. The temperature of the gas is 300 K in each part. Calculate the mass of the piton.

Sol. Let \(l_{1}\) and \(l_{2}\) be the lengths of the upper part and the lower part of the cylinder respectively. clearly, \(l_{1} =\) 50 cm and \(l_{2} = 40\text{ }cm\). Let the perssures in the upper and lower parts be \(p_{1}\) and \(p_{2}\) respectively. Let the area of coss-section of the cylinder be \(A\). The temperature in both parts is \(T = 300\text{ }K\).
Consider the equilibrium of the piston. The forces acting on the pistion are
(a) its weight mg
(b) \(p_{1}\text{ }A\) downward, by the upper part of the gas and (c) \(p_{2}\text{ }A\) upward, by the lower part of the gas.

Thus, \(\ p_{2}\text{ }A = p_{1}\text{ }A + mg\)

Using \(pV = nRT\) for the upper and the lower parts

\[\begin{array}{r} l_{1}\text{ }A = nRT\#(ii) \end{array}\]

and

\[\begin{array}{r} p_{2}l_{2}\text{ }A = nRT.\#(iii) \end{array}\]

Putting \(p_{1}\text{ }A\) and \(p_{2}\text{ }A\) from (ii) and (iii) into (i),

\[\frac{nRT}{l_{2}} = \frac{nRT}{l_{1}} + mg\]

Thus, \(\ m = \frac{nRT}{g}\left\lbrack \frac{1}{l_{1}} - \frac{1}{l_{2}} \right\rbrack\)

\[\begin{matrix} & \ = \frac{(0.1\text{ }mol)(8.3\text{ }J/mol - K)(300\text{ }K)}{9.8\text{ }m/s^{2}}\left\lbrack \frac{1}{0.4\text{ }m} - \frac{1}{0.5\text{ }m} \right\rbrack \\ & \ = 12.7\text{ }kg. \end{matrix}\]

Q. 15 Figure shows a cylindrical tube of volume \(V_{0}\) divided in two parts by a frictionless separator. The walls of the tube are adiabatic but the separator is conducting. Ideal gasses are filled in the two parts. When the separator is kept in the middle, the pressures are \(p_{1}\) and \(p_{2}\) in the left part and the right part respectively. The separator is slowly slide and is released at a position where it can stay in equilibrium. Find the volume of the two parts.

Sol. As the separator is conducting, the temperatures in the two parts will be the same. Suppose the common temperature is T when the separator is in the middle. Let \(n_{1}\) and \(n_{2}\) be the number of moles of the gas in the left part and the right part respectively. Using ideal gas equation,

\[p_{1}\frac{{\text{ }V}_{0}}{2} = n_{1}RT\]

and

\[p_{2}\frac{{\text{ }V}_{0}}{2} = n_{2}RT\]

Thus, \(\ \frac{n_{1}}{n_{2}} = \frac{p_{1}}{p_{2}}\)

The separator will stay in equilibrium at a position where the pressures on the two sides are equal. Suppose the volume of the left part is \(V_{1}\) and of the right part is \(V_{2}\) in this situation. Let the common pressure be p '. Also, let the common temperature in this situation be ' T '.
Using ideal gas equation
and

\[\begin{matrix} & p^{'}V_{1} = n_{1}{RT}^{'} \\ & p^{'}V_{2} = n_{2}{RT}^{'} \end{matrix}\]

or, \(\ \frac{V_{1}}{{\text{ }V}_{2}} = \frac{n_{1}}{n_{2}} = \frac{p_{1}}{p_{2}}\)

Also,

\[\begin{array}{r} V_{1} + V_{2} = V_{0}\#(i) \end{array}\]

Thus,

\[V_{1} = \frac{p_{1}{\text{ }V}_{0}}{p_{1} + p_{2}}\text{~}\text{and}\text{~}V_{2} = \frac{p_{2}{\text{ }V}_{0}}{p_{1} + p_{2}}\]

Q. 16 A gas is heated isobarically and the heat used for external work is W. Find the total amount of heat supplied.
Sol.

\[\begin{matrix} & \Delta Q = \Delta W + \Delta U \\ & \ = nRT + nC_{v}\Delta T = nRT + \frac{nR\Delta T}{\gamma - 1} = (nR\Delta T)\left( \frac{\gamma}{\gamma - 1} \right) \\ & \ = \frac{W\gamma}{\gamma - 1}\ \lbrack\therefore W = nR\Delta T\rbrack \end{matrix}\]

Q. 17 A vessel contains a mixture consisting of \(m_{1} = 7g\) of nitrogen \(\left( M_{1} = 28 \right)\) and \(m_{2} = 11g\) of carbon dioxide \(\left( M_{2} = 44 \right)\) at temperature \(T = 300\text{ }K\) and pressure \(p_{0} = 1\text{ }atm\). Find the density of the mixture.

Sol. Let V be the volume of the vessel. Then \(\rho_{\text{mis}\text{~}} = \frac{m_{1} + m_{2}}{V}\)
Let \(p_{1}\) and \(p_{2}\) be the partial pressure
Then \(p_{1}\text{ }V = \frac{m_{1}}{M_{1}}RT\) and \(p_{2}\text{ }V = \frac{m_{2}}{M_{2}}RT,p_{0} = p_{1} + p_{2}\)

\[\begin{matrix} \therefore & p_{0} = \left( \frac{m_{1}}{M_{1}} + \frac{m_{2}}{M_{2}} \right)\frac{RT}{\text{ }V} \\ \therefore & \rho_{\text{mix}\text{~}} = \frac{\left( m_{1} + m_{2} \right)M_{1}M_{2}}{{\text{ }m}_{1}M_{2} + m_{2}M_{1}} \times \frac{p_{0}}{RT} \\ & = \frac{(7 + 11) \times 28 \times 44 \times 10^{- 3}}{7 \times 44 + 11 \times 28} \times \frac{{10}^{5}}{8.3 \times 300} \\ & = 1.446\text{ }kg{\text{ }m}^{- 3} = 1.446\text{~}\text{per }\text{litre}\text{~} \end{matrix}\]