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Mechanical Properties Of Fluids

Question

Air is expanded from 50 litres to 150 litres at atmosphere the external work done is (1 atmospheric pressure =1 $\times$ $10^{5} N / m^{2}$ )

$2 \times 10^{- 8} J$

$2 \times 10^{4} J$

200J

2000J

Solution

$2 \times 10^{4} J$

$Work done is W = P ∆ V = 2 \times 10^{5} \times (150 - 50) \times 10^{- 3} = 2 \times 10^{4} J$