An ideal gas undergoes a process where its pressure P is related to its volume V by the equation P = kV squared. If the volume changes from V to 2V, what is the work done by the gas?

A. 7/3 kV^3 — The work done is calculated by integrating P dV from V to 2V. Substituting P = kV squared into the integral yields the result 7/3 k V cubed.

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$An ideal gas undergoes a process where its pressure P is related to its volume V by the equation P = k V^{2} . If the volume changes from V to 2 V, what is the work done by the gas?$

A. $7/3 k V^{3}$
B. $8/3 k V^{3}$
C. $5/3 k V^{3}$
D. $2 k V^{3}$

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Correct answer: A — $7/3 k V^{3}$

Solution

The work done is calculated by integrating \int_{V}^{2 V} P d V = \int_{V}^{2 V} k V^{2} d V . Evaluating this integral gives [k V^{3} /3] from V to 2 V, which equals k (8 V^{3} - V^{3}) /3 = 7/3 k V^{3} .

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An ideal gas undergoes a process where its pressure P is related to its volume V by the equation P=kV2. If the volume changes from V to 2V, what is the work done by the gas?

Solution

Attempt this question & track your score

More Thermodynamics practice questions

$An ideal gas undergoes a process where its pressure P is related to its volume V by the equation P = k V^{2} . If the volume changes from V to 2 V, what is the work done by the gas?$