Question
A gas is expanded from volume $${V_0}$$ to $$2{V_0}$$ under three different processes. Process 1 is isobaric process, process 2 is isothermal and process 3 is adiabatic.
A gas is expanded from volume $${V_0}$$ to $$2{V_0}$$ under three different processes. Process 1 is isobaric process, process 2 is isothermal and process 3 is adiabatic.
Let $$\Delta {U_1},\Delta {U_2}$$ and $$\Delta {U_3}$$ be the change in internal energy of the gas is 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.
$$\Delta {U_2} < \Delta {U_3} < \Delta {U_1}$$
Answer :
$$\Delta {U_1} > \Delta {U_2} > \Delta {U_3}$$
Solution :
Since volume is same in all three process therefore temperature will be least having least pressure.
Since volume is same in all three process therefore temperature will be least having least pressure.
