1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

A sample of an ideal gas has pressure p0, volume V0 and temperature T0. It is isothermally expanded to twice its original volume. It is then compressed at constant pressure to the original volume V0. Finally the gas is heated at constant volume to get the original temperature.

A
Heat absorbed in the process is p0V0 ln2.
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
Work done in the process is p0V0(ln212).
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
C
Internal energy doesn't change.
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
D
Heat absorbed in the process is p0V0(ln212).
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
Open in App
Solution

The correct options are B Work done in the process is p0V0(ln2−12). C Internal energy doesn't change. D Heat absorbed in the process is p0V0(ln2−12).The process is cyclic so that the change in internal energy is zero. The heat supplied is therefore equal to the work done by the gas. The work done during step ab is W1=nRT0ln2=p0V0ln2 Also from the ideal gas equation, paVa=pbVb ⇒pb=paVaVb =p0V02V0=p02 In the step bc, pressure remains constant. Hence, the work done is W2=pb(Vc−Vb)=p02(V0−2V0)=−p0V02 In the step ca, volume remains constant. Therefore work done is zero. ∴ Net work done by the gas in the process is W=W1+W2=p0V0(ln2−12) Hence, the heat supplied to the gas is also p0V0(ln2−12)

Suggest Corrections
2
Join BYJU'S Learning Program
Related Videos
Boyle's Law
CHEMISTRY
Watch in App
Explore more
Join BYJU'S Learning Program