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(D) The internal energy of an ideal gas is given by $U = \frac{f}{2} nRT$,where $f$ is the degrees of freedom. For a monoatomic gas,$f=3$,so $U \propt

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$U_B - U_A < 0$

Vedclass · Answer

(D) The internal energy of an ideal gas is given by $U = \frac{f}{2} nRT$,where $f$ is the degrees of freedom. For a monoatomic gas,$f=3$,so $U \propto T$.From the graph:State $A$: $T_A = T$State $B$: $T_B = 2T$State $C$: $T_C = 2T$State $D$: $T_D = T$Comparing internal energies:$U_A = U_D$ (since $T_A = T_D$),so $U_A - U_D = 0$. (True)$U_B = U_C$ (since $T_B = T_C$),so $U_B - U_C = 0$. (True)$U_C > U_D$ (since $T_C > T_D$),so $U_C - U_D > 0$. (True)$U_B > U_A$ (since $T_B > T_A$),so $U_B - U_A > 0$. Therefore,$U_B - U_A < 0$ is false.

An ideal monoatomic gas is taken through the thermodynamic states $A \to B \to C \to D$ via the paths shown in the figure. If $U_A, U_B, U_C$ and $U_D$ represent the internal energy of the gas in states $A, B, C$ and $D$ respectively,then which of the following is not true?

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