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Work

the volume of...

Question

the volume of a gas expands by 0.25 $m^{3}$ at a constant pressure of $10^{3} N / m^{2}$ . The work done is equal to

2.5 erg

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250 j

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250 w

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250 N

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Solution

The correct option is B 250 j
Given,
$d V = 0.25 m^{3}$
$P = 10^{3} N / m^{2}$
Work done, $W = \int P d V$
$W = 0.25 \times 1000 = 250 J$
The correct option is B.

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A gas expands $0.25 m^{3}$ at constant pressure $10^{3} N / m^{2}$ , the work done is

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1 \times 10^{5} {N m}^{- 2}

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