Question

Calculate magnitude of minimum work done for ideal gas when change of state take place from (3 atm, 1lit) to (1 atm, 3 lit) at 300 k

Answer 1

Magnitude of minimum work done for ideal gas is -2.757 kJ.

Given:

Temperature = T = 300 K

Pressure = P1 = 3 atm

Volume = V1 = 1 L

Pressure = P2 = 1 atm

Volume = V2 = 3 L

To find:

Work done = ?

Formula used:

$\bold{w=-nRT \ln[\frac{V2}{V1}]}$

Where,

$\bold{R=8.314 \ J/mol.K}$

Solution:

For minimum work done, n = 1

$\bold{w=1\times8.314\times300\times2.303\log[\frac{3}{1}]}$

$\bold{w=-5744.14\times \log3}$

$\bold{w=-5744.14\times 0.48}$

$\bold{w=-2757.28 \ J}$

$\bold{\therefore w= -2.757 \ kJ}$