Question

A 1 mole sample of an ideal gas is kept at 0.0°C during an expansion from 3.0L to 10.0L.
How much work is done by the gas during the expansion (8.3J/K.moll)? ​

Answer 1

Given :

•Number of moles =1

•Temperatures =0.0°C

• V1=3.0L •V2=10.0L

•R=8.314 J/k.mole

To Find :

• How much work is done by the gas during the expansion =?

Formula :

• The work done by the gas during the expansion is

$\\ \blue\bigstar\boxed{\sf{W=nRTlog\bigg(\dfrac{V_{2}}{V_{1}}\bigg) }}$

Solution :

• As, the temperature of gas is fixed .This process is Isothermal.

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

Substituting the given values :

$\\ \implies\sf{W=(1) \times (8.314) \times (273) log\bigg(\dfrac{3}{10}\bigg) }$

$\\ \implies\sf{W=1 \times 8.314 \times 273 log\bigg(\dfrac{3}{10} \bigg) }$

$\\ \implies\sf{ W=-2.7 \times 10^{3} J}$

Hence, The work done by the gas during the expansion is W=-2.7 x 10³ J.