MHT CET · Physics · Thermodynamics
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
- A increased by \(400 \mathrm{~J}\)
- B increased by \(200 \mathrm{~J}\)
- C increased by \(100 \mathrm{~J}\)
- D decreased by \(200 \mathrm{~J}\)
Answer & Solution
Correct Answer
(A) increased by \(400 \mathrm{~J}\)
Step-by-step Solution
Detailed explanation
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.
\(\therefore \Delta \mathrm{U}=\mathrm{Q}-\mathrm{P} \Delta \mathrm{V}=100-[50 \times(-6)]=400 \mathrm{~J}\)
As, \(\Delta \mathrm{U}\) is positive, the internal energy is increased.
\(\mathrm{Q}=\Delta \mathrm{U}+\Delta \mathrm{W}=\Delta \mathrm{U}+\mathrm{P} \Delta \mathrm{V}\)
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.
\(\therefore \Delta \mathrm{U}=\mathrm{Q}-\mathrm{P} \Delta \mathrm{V}=100-[50 \times(-6)]=400 \mathrm{~J}\)
As, \(\Delta \mathrm{U}\) is positive, the internal energy is increased.
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