Question

The internal energy of a gas is given by $U=1.5\text{pv}$. It expands from 100 $c{m}^3$ to 200 $c{m}^3$ against a constant pressure of $1.0\times {10}^5\text{Pa}$. Calculate the heat absorbed by the gas in the process.

Answer 1

$\Delta \text{U}=1.5\text{pV}$

$\Delta \text{V}=(200-100){\text{cm}}^3=100{\text{cm}}^3$

$={10}^{-4}{\text{m}}^3$

$P=1\times {10}^5\text{Pa}$

$\Delta \text{U}=1.5\times {10}^5\times {10}^{-4}=15\text{J}$

$\Delta \text{W}={10}^5\times {10}^{-4}=10\text{J}$

$\Delta \text{Q}=\Delta \text{U}+\Delta \text{W}=10+15=25\text{J}$