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A gas expands from 10litre to 20litre at a constant pressure of 2 atmospheres. What is the work done by the gas during the expansion. (1atm $= 1.013 \times 10^{5} P a)$

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$2.026 \times 10^{2} J$

$2.026 \times 10^{1} J$

$2.026 \times 10^{3} J$

$2.026 \times 10^{4} J$

answer is C.

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Detailed Solution

$P = 2 a t m = 2 \times 1.013 \times 10^{5} = 2.026 \times 10^{5} N / m^{2}$

$V_{1} = 10 l t = 10 \times 10^{3} c m^{3} = 10^{- 2} m^{2}$

$V_{2} = 20 l t = 2 \times 10^{- 2} m^{3}$

Work done

$= P (V_{2} - V_{1}) = 2.026 \times 10^{5} (2 \times 10^{- 2} - 1 \times 10^{- 2})$

$= 2.026 \times 10^{3} J$

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