JEE Advance - Chemistry (2021 - Paper 2 Online - No. 17)
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 $$\ln {{{V_3}} \over {{V_2}}}$$ is _________.
(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 $${5 \over 2}$$R)
(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 $${5 \over 2}$$R)
Answer
10
Explanation
For process 1, entropy is constant thus, q is constant
$$\therefore$$ $${W_I} = \Delta U = n{C_{V,m}}\Delta T$$
$$ = - (2250 - 450)R = - 1800R$$ .... (i)
$$\Delta U = n{C_{V,m}}\Delta T$$
$$ - 1800R = 1 \times {{5R} \over 2} \times \Delta T$$
$$ \Rightarrow \Delta T = - 720K$$
$${T_2} - {T_1} = - 720K$$
$${T_2} = - 720K + 900K = 180K$$
For process II,
$${W_{II}} = - nR{T_2}\ln \left( {{{{V_3}} \over {{V_2}}}} \right)$$
$$ = - 1 \times R \times 180\ln \left( {{{{V_3}} \over {{V_2}}}} \right)$$ .... (ii)
As, both work done for process I and II are equal,
Therefore, WI = WII
$$ - 1800R = - R \times 180 \times \ln \left( {{{{V_3}} \over {{V_2}}}} \right)$$
$$\ln \left( {{{{V_3}} \over {{V_2}}}} \right) = {{1800} \over {180}} \Rightarrow \ln \left( {{{{V_3}} \over {{V_2}}}} \right) = 10$$
$$\therefore$$ $${W_I} = \Delta U = n{C_{V,m}}\Delta T$$
$$ = - (2250 - 450)R = - 1800R$$ .... (i)
$$\Delta U = n{C_{V,m}}\Delta T$$
$$ - 1800R = 1 \times {{5R} \over 2} \times \Delta T$$
$$ \Rightarrow \Delta T = - 720K$$
$${T_2} - {T_1} = - 720K$$
$${T_2} = - 720K + 900K = 180K$$
For process II,
$${W_{II}} = - nR{T_2}\ln \left( {{{{V_3}} \over {{V_2}}}} \right)$$
$$ = - 1 \times R \times 180\ln \left( {{{{V_3}} \over {{V_2}}}} \right)$$ .... (ii)
As, both work done for process I and II are equal,
Therefore, WI = WII
$$ - 1800R = - R \times 180 \times \ln \left( {{{{V_3}} \over {{V_2}}}} \right)$$
$$\ln \left( {{{{V_3}} \over {{V_2}}}} \right) = {{1800} \over {180}} \Rightarrow \ln \left( {{{{V_3}} \over {{V_2}}}} \right) = 10$$
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