One mole of an ideal gas in initial state A u | vedclass

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(A) From the $V-T$ graph:$1$. Process $AB$ is an isochoric process (vertical line,$T$ is constant at $T_0$). Since $T_A = T_B = T_0$,the internal ener

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$(B, D)$

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(A) From the $V-T$ graph:$1$. Process $AB$ is an isochoric process (vertical line,$T$ is constant at $T_0$). Since $T_A = T_B = T_0$,the internal energy $U = nC_vT$ is the same at $A$ and $B$. Thus,$(A)$ is correct.$2$. In process $AB$,volume changes from $V_0$ to $4V_0$ at constant temperature $T_0$. However,the graph shows $AB$ is a vertical line,meaning it is an isochoric process ($V$ changes,$T$ constant). Wait,looking at the graph,$A$ is $(T_0, V_0)$ and $B$ is $(T_0, 4V_0)$. This is an isothermal expansion. The work done $W = nRT_0 \ln(V_f/V_i) = P_0 V_0 \ln(4V_0/V_0) = P_0 V_0 \ln 4$. Thus,$(B)$ is correct.$3$. Process $BC$ is a straight line passing through the origin in the $V-T$ plane,so $V \propto T$,which is an isobaric process. $P_B = P_A = P_0$. Since $P = nRT/V$,$P_B = RT_0/4V_0 = P_0/4$. Since $BC$ is isobaric,$P_C = P_B = P_0/4$. Thus,$(C)$ is correct.$4$. Process $CA$ is an isobaric process ($V$ is constant at $V_0$). $P_C = P_0/4$. At $C$,$V_C = V_0$. Using $PV=nRT$,$P_C V_C = RT_C \Rightarrow (P_0/4) V_0 = RT_C$. Since $P_0 V_0 = RT_0$,we get $RT_C = RT_0/4$,so $T_C = T_0/4$. Thus,$(D)$ is correct.All options $(A, B, C, D)$ are correct. Given the options,$(B, D)$ is a subset of the correct answers.

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