A gas at a pressure of 10 5 Nm -2 expands from 0.6m -3 to 1.2m 3 at constant temperature, the work done is

Correct Answer is : 6.0 x 10 4J

Correct Answer is : 6.0 x 10 4J

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JAMB Physics 2013 Paper

Question : 14

Total: 43

Solution:

Concept:

When a gas expands and does work against a constant pressure, the work done is the product of the pressure and the increase in volume:

W = P\Delta V

Explanation:

Pressure,

P = 10^{5} \, \mathrm{N\,m^{-2}}

Initial volume,

V_{1} = 0.6 \, \mathrm{m^{3}}

Final volume,

V_{2} = 1.2 \, \mathrm{m^{3}}

The change in volume is

\Delta V = V_{2} - V_{1} = 1.2 - 0.6 = 0.6 \, \mathrm{m^{3}}

Substitute into the work formula:

W = P\Delta V = 10^{5} \times 0.6

So,

W = 6.0 \times 10^{4} \, \mathrm{N\,m}

Since

1 \, \mathrm{N\,m} = 1 \, \mathrm{J}

, we have

W = 6.0 \times 10^{4} \, \mathrm{J}

Answer:

The work done is

6.0 \times 10^{4} \, \mathrm{J}

; the correct option is D.

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