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Question

Work done is equal to area under the P-V graph, thus work done by gas is greater for isothermal process.

For isothermal process: $T_2=T_0$

For a

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Work done is equal to area under the P-V graph, thus work done by gas is greater for isothermal process.

For isothermal process: $T_2=T_0$

For adibatic process: $T_0\left(V_0\right)^{\gamma-1}=T(V)^{\gamma-1}$

As $V > V _0 \Rightarrow T < T _0$

Thus final temp. is greater for isothermal process.

For isothermal process: $P_0 V_0=P_i V \Rightarrow P_1=P_0 \frac{V_0}{V}$

For adibatic process: $P_0\left( V _0\right)^\gamma= P _2( V )^y$

$\Rightarrow P _2= P _0\left(\frac{ V _0}{ V }\right)^\gamma$

As $\gamma > 1$ (always) $\Rightarrow P_1 > P_2$

Column$-I$	Column$-II$
$(a)$ Adiabatic	$(i)$ $\Delta Q = \Delta U$
$(b)$ Isothermal	$(ii)$ $\Delta Q = \Delta W$
	$(iii)$ $\Delta U = -\Delta W$

Two gases have the same initial pressure, volume and temperatue. They expand to the same final volume, one adiabatically and the other isothermally, if the two gases are compressed to the same final volume

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