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(A) For an ideal gas undergoing compression from $2V_0$ to $V_0$:$1$. Reversible Isothermal: $T_f = T_i$. Work is done on the system,but heat is relea

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Final temperature of gas will be highest at the end of second process.

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(A) For an ideal gas undergoing compression from $2V_0$ to $V_0$:$1$. Reversible Isothermal: $T_f = T_i$. Work is done on the system,but heat is released to maintain constant temperature.$2$. Reversible Adiabatic: $T_f V_f^{\gamma-1} = T_i V_i^{\gamma-1}$. Since $V_f  T_i$.$3$. Irreversible Adiabatic: The work done is $W = -P_{ext} \Delta V$. Since the process is adiabatic $(q=0)$,$\Delta U = W$. The temperature rise is lower than the reversible adiabatic case because the work done is less efficient.Comparing the final temperatures: $T_{rev, adiabatic} > T_{irrev, adiabatic} > T_{isothermal}$.Thus,the final temperature is highest at the end of the second process.

List-$I$	List-$II$
$A$. At constant volume the change in internal energy of a system	$I$. $W = -2.303 nRT \log \frac{V_f}{V_i}$
$B$. Isothermal irreversible change	$II$. $W_{adiabatic} = \Delta U$
$C$. Isothermal reversible change	$III$. $q_V = \Delta U$
$D$. Adiabatic change	$IV$. $W = -p_{ex} (V_f - V_i)$
	$V$. $\Delta U = \Delta H - \Delta nRT$

$A$ sample of gas is compressed from an initial volume of $2V_0$ to $V_0$ using three different processes.
First: Using reversible isothermal.
Second: Using reversible adiabatic.
Third: Using irreversible adiabatic under a constant external pressure.
Then

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$A$ sample of gas is compressed from an initial volume of $2V_0$ to $V_0$ using three different processes.First: Using reversible isothermal.Second: Using reversible adiabatic.Third: Using irreversible adiabatic under a constant external pressure.Then