Question

A gas is expanded from volume $V_0$ to $2V_0$ under three different processes,as shown in the figure. Process 1 isobaric process, process 2 is isothermal and process 3 is adiabatic.
Let $\Delta U_1$, $\Delta U_2$, $\Delta U_3$ be the change in internal energy of the gas in these three processes, then :

• A. $\Delta U_1$ > $\Delta U_2$ > $\Delta U_3$
• B. $\Delta U_1$ < $\Delta U_2$ < $\Delta U_3$
• C. $\Delta U_2$ < $\Delta U_1$ < $\Delta U_3$
• D. $\Delta U_3$ < $\Delta U_3$ < $\Delta U_1$

Answer 1

The correct option is A $\Delta U_1$ > $\Delta U_2$ > $\Delta U_3$

From first law of thermodynamics,

Heat supplied $\left(\Delta Q\right)=$ Work done by system $\left(\Delta W\right)+$ Increase in internal energy $\left(\Delta U\right)$

All three processes being expansions,

$\Delta W>0$ , for each of them

For isothermal process, $\Delta U=0$

For adiabatic process, $\Delta Q=0$

For isobaric expansion,$\Delta U=n{C}_p\Delta T>0$

Thus, for process 1, $\Delta U_1>0$

For process $2,\Delta U_2=0$

For process $3,\Delta U_3=-\Delta W_3<0$

Thus, $\Delta U_1>\Delta U_2>\Delta U_3$