The equation of state for a gas is given by P | vedclass

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Question

(A) Given the equation of state: $PV = nRT + \alpha V$.For $n = 1$ mole,$PV = RT + \alpha V$,which can be rewritten as $V(P - \alpha) = RT$,or $V = \f

Vedclass · Accepted Answer

$\frac{P_0 T_0 R}{P_0 - \alpha}$

Vedclass · Answer

(A) Given the equation of state: $PV = nRT + \alpha V$.For $n = 1$ mole,$PV = RT + \alpha V$,which can be rewritten as $V(P - \alpha) = RT$,or $V = \frac{RT}{P - \alpha}$.Initially,at $T = T_0$ and $P = P_0$,the volume is $V_0 = \frac{RT_0}{P_0 - \alpha}$.Since the process is isobaric,$P$ remains constant at $P_0$. When the temperature doubles,$T_f = 2T_0$.The final volume is $V_f = \frac{R(2T_0)}{P_0 - \alpha} = 2V_0$.The work done in an isobaric process is $W = P_0(V_f - V_0)$.Substituting the values: $W = P_0(2V_0 - V_0) = P_0 V_0$.Since $V_0 = \frac{RT_0}{P_0 - \alpha}$,we have $W = P_0 \left( \frac{RT_0}{P_0 - \alpha} \right) = \frac{P_0 T_0 R}{P_0 - \alpha}$.

Column-$I$	Column-$II$
$(a)$ Graph $(a)$	$(i)$ Adiabatic process
$(b)$ Graph $(b)$	$(ii)$ Isobaric process
	$(iii)$ Isochoric process

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