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One mole of an ideal monoatomic gas is taken from A to C along the path A B C. The temperature of the gas at A is T_{0} For the process A B C -

Option: 1

work done by the gas is R T_{0}


Option: 2

work done by the gas is 5RT_0


Option: 3

work done by the gas is 10RT_0


Option: 4

None of these


Answers (1)

best_answer

Here, n=1

in process A B ; p= constant

\therefore \vee \propto T

\text { As } V_{B}=2 V_{A} \quad \therefore T_{B}=2 T_{A}=2 T_{0}

\begin{aligned} \therefore T_{0} R & =P_{0} V_{0}=\Delta \omega_{A B} \\ & \Delta Q_{A B}=C_{P} \Delta t=\frac{5}{2} R\left(2 T_{0}-T_{0}\right)=\frac{5}{2} R T_{0} \\ \therefore & \Delta U_{A B}=\Delta Q_{A B}-\Delta \omega_{A B}=\frac{5}{2} R T_{0}-R T_{0}=\frac{3}{2} R T_{0} \end{aligned}

In process BC,  v= constant \therefore P \alpha T

\begin{aligned} & P_{C}=2 P_{B} \quad \therefore T_{C}=2 T_{B}=4 T_{0} \\ & \Delta \omega_{B C}=0, \Delta Q_{B C}=\Delta U_{B C}=C_{V}(\Delta T) \\ & =\frac{3}{2} R\left(4 T_{0}-2 T_{0}\right)=3 R T_{0} \\ & \Delta \omega_{\text {net }}=R T_{0}, \Delta Q_{\text {net }}=\frac{11}{2} R T_{0} \\ & \text { and } \Delta U_{\text {net }}=\frac{9}{2} R T_{0} \end{aligned}

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jitender.kumar

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