The work done on the system when one mole of an ideal gas at 500 K is compressed isothermally and reversibly to 1 / 10 th of its original volume ( R =2 cal ) -

Question

The work done on the system when one mole of an ideal gas at 500 K is compressed isothermally and reversibly to 1 / 10 th of its original volume ( R =2 cal ) - (A) 500 kCal (B) 1.51 kCal (C) -23.03 kCal (D) 2.303 kCal

Tardigrade Admin · Accepted Answer

w=-2.303 R T log (v2/v1) =-2.303 × 2 × 500 log (1/10) =-2.303 × 103 log 10-1 =2.303 kCal

Q. The work done on the system when one mole of an ideal gas at $500 K$ is compressed isothermally and reversibly to $1/10$ th of its original volume $(R = 2$ cal $)$ -

500 kCal

1.51 kCal

-23.03 kCal

2.303 kCal

$w = - 2.303 RT lo g \frac{v _{2}}{v _{1}}$

$= - 2.303 \times 2 \times 500 lo g \frac{1}{10}$

$= - 2.303 \times 1 0^{3} lo g 1 0^{- 1}$

$= 2.303 k C a l$

Q. The work done on the system when one mole of an ideal gas at 500K is compressed isothermally and reversibly to 1/10 th of its original volume (R=2 cal ) -