Question

A cyclic process for 1 mole of an ideal gas is shown in the V-T diagram. The work done in AB, BC and CA respectively are

• A. 0, RT1ln $\left( \frac { V _ { 1 } } { V _ { 2 } } \right)$ , R(T1 – T2)
• B. R, (T1 – T2)R, RT1ln $\left( \frac { V _ { 1 } } { V _ { 2 } } \right)$
• C. 0, RT2ln, (V1 – V2)
• D. 0, RT2ln $\left( \frac { V _ { 1 } } { V _ { 2 } } \right)$ , R(T1 – T2)

Answer 1

Answer: (C)

0, RT2ln, (V1 – V2)

Answer 2

:

During AB, process is isochoric.

$\therefore \Delta V = 0 \therefore W = 0$

During BC, process is isothermal

$\therefore \Delta T = 0$ $\therefore W = R T _ { 2 } \ln \frac { V _ { 2 } } { V _ { 1 } }$

During CA, process is isobaric, so pressure is

constant $\therefore W = P \left( V _ { 1 } - V _ { 2 } \right)$

But $P V _ { 1 } = R T _ { 1 }$

$\therefore P = \frac { R T _ { 1 } } { V _ { 1 } } = \frac { R T _ { 2 } } { V _ { 2 } }$ $\therefore W = \frac { R T _ { 1 } } { V _ { 1 } } \left( V _ { 1 } - V _ { 2 } \right) = \frac { R T _ { 2 } } { V _ { 2 } } \left( V _ { 1 } - V _ { 2 } \right)$