One mole of an ideal gas is taken from A to B | vedclass

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(A) For an ideal gas,the internal energy $U$ is a function of temperature only,given by $U = nC_{v}T$.From the given $V-T$ graph,the process $A \right

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at $A$ and $B$ are equal

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(A) For an ideal gas,the internal energy $U$ is a function of temperature only,given by $U = nC_{v}T$.From the given $V-T$ graph,the process $A \rightarrow B$ is a vertical line,which means the temperature $T$ is constant. Thus,$A \rightarrow B$ is an isothermal process.Since the temperature at $A$ $(T_{A})$ is equal to the temperature at $B$ $(T_{B})$,the internal energy at $A$ is equal to the internal energy at $B$.For the process $B \rightarrow C$,the volume $V$ is constant,and the temperature increases from $T_{B}$ to $T_{C}$. Since $T_{C} > T_{B}$,the internal energy at $C$ is greater than at $B$.Therefore,the correct statement is that the internal energy at $A$ and $B$ are equal.

One mole of an ideal gas is taken from $A$ to $B$,from $B$ to $C$,and then back to $A$. The variation of its volume with temperature for that change is as shown. Its pressure at $A$ is $P_{0}$ and volume is $V_{0}$. Then,the internal energy:

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