An ideal gas at pressure P and volume V is ex | vedclass

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(A) For an ideal gas, initial state is $(P, V)$ and final volume is $V_f = 2V$.$(p)$ Isobaric process: Pressure $P$ is constant.Work $W = P \Delta V =

Vedclass · Accepted Answer

$p-y, q-z, r-x$

Vedclass · Answer

(A) For an ideal gas, initial state is $(P, V)$ and final volume is $V_f = 2V$.$(p)$ Isobaric process: Pressure $P$ is constant.Work $W = P \Delta V = P(2V - V) = PV$. Thus, $(p) ightarrow (y)$.$(q)$ Isothermal process: Temperature $T$ is constant.Work $W = nRT \ln(\frac{V_f}{V_i}) = PV \ln(\frac{2V}{V}) = PV \ln 2$. Thus, $(q) ightarrow (z)$.$(r)$ Adiabatic process: $PV^{\gamma} = 	ext{constant}$.$P_i V_i^{\gamma} = P_f V_f^{\gamma} \Rightarrow P(V)^{\gamma} = P_f(2V)^{\gamma} \Rightarrow P_f = P(2)^{-\gamma}$.Work $W = \frac{P_i V_i - P_f V_f}{1 - \gamma} = \frac{PV - (P 2^{-\gamma})(2V)}{1 - \gamma} = \frac{PV(1 - 2^{1 - \gamma})}{1 - \gamma} = \frac{PV(2^{1 - \gamma} - 1)}{\gamma - 1}$.Wait, checking the sign: $W = \frac{P_i V_i - P_f V_f}{\gamma - 1} = \frac{PV - P 2^{-\gamma} 2V}{\gamma - 1} = \frac{PV(1 - 2^{1 - \gamma})}{\gamma - 1}$. Thus, $(r) ightarrow (x)$.Therefore, the correct matching is $p-y, q-z, r-x$.

Column $I$	Column $II$
$(p)$ isobaric	$(x)$ $\frac{PV(1 - 2^{1 - \gamma})}{\gamma - 1}$
$(q)$ isothermal	$(y)$ $PV$
$(r)$ adiabatic	$(z)$ $PV \ln 2$

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