Department of Pre-University Education, KarnatakaPUC Karnataka Science Class 11
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Figure Shows a Cylindrical Tube of Volume V with Adiabatic Walls Containing an Ideal Gas. the Internal Energy of this Ideal Gas is Given by 1.5 Nrt. - Physics

Sum

Figure shows a cylindrical tube of volume V with adiabatic walls containing an ideal gas. The internal energy of this ideal gas is given by 1.5 nRT. The tube is divided into two equal parts by a fixed diathermic wall. Initially, the pressure and the temperature are p1, T1 on the left and p2, T2 on the right. The system is left for sufficient time so that the temperature becomes equal on the two sides. (a) How much work has been done by the gas on the left part? (b) Find the final pressures on the two sides. (c) Find the final equilibrium temperature. (d) How much heat has flown from the gas on the right to the gas on the left?

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Solution

Let n1 , U1 and n2 ,U2 be the no. of moles , internal energy of ideal gas in the left chamber and right chamber respectively.

(a) As the diathermic wall is fixed, so final volume of the chambers will be same. Thus, ΔV = 0, hence work done ΔW= P Δ V = 0

by eq. of state in the first and second chamber

`P_1V/2=n_1RT_1`

`rArr n_1=(P_1V)/(2RT_1)`

`P_2V/2=n_2RT_2`

`rArr n_2=(P_2V)/(2RT_2)`

n = n1 + n2

`rArrn=(P_1V)/(2RT_1)+(P_2V)/(2RT_2)=V/(2R)((P_1T_2+P_2T_1)/(T_1T_2))`

Again,

U = nCvT

`rArrnC_"v"T=1.5nRT`

`rArrC_"v"=1.5R`

In the first and second chamber internal energy is given by

U1 = n1CvT1 = n1 1.5RT1

U2 = n2CvT2 = n2 1.5RT2

U = U1 + U2

1.5nRT = n1 1.5RT1 + n2 1.5RT2

⇒ nT = n1T1 + n2T2

`rArrnT = (P_1V)/(2RT_1)T_1+(P-2V)/(2RT_2)T_2=((P_1+P_2)V)/(2R)`

`rArrT=((P_1+P_2)V)/(2nR)=((P_1+P_2)V)/(2RV/(2R)((P_1T_2+P_2T_1)/(T_1T_2)))=(T_1T_2(P_1+P_2))/(P_1T_2+P_2T_1)................(1)`

b) Let final pressure in the first and second compartment P1' and P2'.

By five variable equ of state in the first chamber

`(P_1V/2)/T_1=(P_1'V/2)/T`

`rArrP_1'=P_1/T_1T`

By eq. (1)

`rArrP_1'=P_1/T_1(T_1T_2(P_1+P_2))/(P_1T_2+P_2T_1)=(P_1T_2(P_1+P_2))/(P_1T_2+P_2T_1)`

Similarly,

`rArrP_1'=P_2/T_2T=(P_2T_1(P_1+P_2))/(P_1T_2+P_2T_1)`

c) Final temperature will be

`T=(T_1T_2(P_1+P_2))/(P_1T_2+P_2T_1).............\left(\text{by equation (1)}\right)`

d) Heat lost by right chamber will be

n2CvT2 - n2CvT

`=(P_2V)/(2RT_2)1.5RT_2-(P_2V)/(2RT_2)1.5R(T_1T_2(P_1+P_2))/(P_1T_2+P_2T_1)`

`=(3P_2V)/4-(3P_2V)/4(T_1(P_1+P_2))/(P_1T_2+P_2T_1)`

`=(3P_2V)/4[1-(T_1(P_1+P_2))/(P_1T_2+P_2T_1)]`

`=(3P_2V)/4[(P_1T_2+P_2T_1-T_1(P_1+P_2))/(P_1T_2+P_2T_1)]`

`=(3P_1P_2V)/4[(T_2-T_1)/(P_1T_2+P_2T_1)]`

Concept: Heat, Internal Energy and Work
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HC Verma Class 11, Class 12 Concepts of Physics Vol. 2
Chapter 4 Laws of Thermodynamics
Q 21 | Page 64
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