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QJEEAdv21Chf022: JEE Advanced 2021 Chemistry - Class 11 JEEAdv21Chf022 Solution: One mole of an ideal gas at 900 K, undergoes two reversible…

One mole of an ideal gas at 900 K, undergoes two reversible processes, I followed by II, as shown below. If the work done by the gas in the two processes are same, the value of ln⁡V3V2\ln {{{V_3}} \over {{V_2}}} is _________.

JEE Advanced 2021 Paper 2 Online Chemistry - Thermodynamics Question 17 English
(U : internal energy, S : entropy, p : pressure, V : volume, R : gas constant)

(Given : molar heat capacity at constant volume, CV,m of the gas is 52{5 \over 2} R)

Step-by-Step Verified Solution

For process 1, entropy is constant thus, q is constant

∴\therefore WI=ΔU=nCV,mΔT{W_I} = \Delta U = n{C_{V,m}}\Delta T

=−(2250−450)R=−1800R= - (2250 - 450)R = - 1800R .... (i)

ΔU=nCV,mΔT\Delta U = n{C_{V,m}}\Delta T

−1800R=1×5R2×ΔT- 1800R = 1 \times {{5R} \over 2} \times \Delta T

⇒ΔT=−720K\Rightarrow \Delta T = - 720K

T2−T1=−720K{T_2} - {T_1} = - 720K

T2=−720K+900K=180K{T_2} = - 720K + 900K = 180K

For process II,

WII=−nRT2ln⁡(V3V2){W_{II}} = - nR{T_2}\ln \left( {{{{V_3}} \over {{V_2}}}} \right)

=−1×R×180ln⁡(V3V2)= - 1 \times R \times 180\ln \left( {{{{V_3}} \over {{V_2}}}} \right) .... (ii)

As, both work done for process I and II are equal,

Therefore, WI = WII

−1800R=−R×180×ln⁡(V3V2)- 1800R = - R \times 180 \times \ln \left( {{{{V_3}} \over {{V_2}}}} \right)

ln⁡(V3V2)=1800180⇒ln⁡(V3V2)=10\ln \left( {{{{V_3}} \over {{V_2}}}} \right) = {{1800} \over {180}} \Rightarrow \ln \left( {{{{V_3}} \over {{V_2}}}} \right) = 10

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