Question

10 moles
of O2 gas are allowed to expand
Isothermally (T = 300k from an initial volume of
10 liters to a final volume of 30 liters . what is the
work done​

Answer 1

Answer:

The work done W=−2.303nRTlog

10

V

1

V

2

W=−2.303× 1 mol × 8.314 J/mol/K × 300 K×log

10

5 m

3

10 m

3

W=− 1729 J

The work done in the expansion of the gas is − 1729 J.

Answer 2

Given:

10 moles of O2 gas are allowed to expand

isothermally (T = 300k from an initial volume of

10 liters to a final volume of 30 liters .

To find:

Work done in this thermodynamic process ?

Calculation:

General expression for work done in isothermal expansion :

$\boxed{ \bold{W = nRT \: ln( \dfrac{V_{2}}{V_{1}} ) }}$

• n = number of moles
• R = Universal Gas constant
• T = Temperature
• V_(2) = final volume, V_(1) = initial volume

Putting available values in SI units:

$\implies \sf W = nRT \: ln( \dfrac{V_{2}}{V_{1}} )$

$\implies \sf W = 2.303 \times nRT \: log ( \dfrac{V_{2}}{V_{1}} )$

$\implies \sf W = 2.303 \times 10 \times R \times 300 \times \: log ( \dfrac{30}{10} )$

$\implies \sf W = 2.303 \times 10 \times R \times 300 \times \: log ( 3)$

$\implies \sf W = 2.303 \times R \times 3000 \times \: log ( 3)$

$\implies \sf W = 2303 \times R \times 3 \times \: log ( 3)$

$\implies \sf W = 6909 \times R \times \: log ( 3)$

$\implies \sf W = 6909 \times 8.314 \times \: log ( 3)$

$\implies \sf W = 27406.52 \: joule$

$\implies \sf W = 27.4 \:kilo joule$

So, work done is 27.4 kJ.