A gas is compressed at a constant pressure of $50 \mathrm{~N} / \mathrm{m}^2$ from a volume of $10…
A gas is compressed at a constant pressure of $50 \mathrm{~N} / \mathrm{m}^2$ from a volume of $10 \mathrm{~m}^3$ to a volume of $4 \mathrm{~m}^3$. Energy of $100 \mathrm{~J}$ is then added to the gas by heating. Its internal energy is
increased by $400 \mathrm{~J}$
increased by $200 \mathrm{~J}$
increased by $100 \mathrm{~J}$
decreased by $200 \mathrm{~J}$
Solution
From first law of thermodynamics,
$\mathrm{Q}=\Delta \mathrm{U}+\Delta \mathrm{W}=\Delta \mathrm{U}+\mathrm{P} \Delta \mathrm{V}$
Change in yolume due to compression
$\Delta \mathrm{V}=\mathrm{V}_2-\mathrm{V}_1=4-10=-6 \mathrm{~m}^3$
Negatiye sign indicates gas is compressed.
$\therefore \quad \Delta \mathrm{U}=\mathrm{Q}-\mathrm{P} \Delta \mathrm{V}=100-[50 \times(-6)]=400 \mathrm{~J}$
As, $\Delta \mathrm{U}$ is positive, the internal energy is increased.
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