An ideal gas expands from $10^{- 3} m^{3}$ to $10^{- 2} m^{3}$ at $300$ K against a constant pressure of $10^{5} N m^{- 2}$ . The work done is?

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Solution

Let us consider the problem:
$W = - P Δ V$
For
$P = 10^{5} N m^{- 2} - 1$ atm
Implies that
$Δ V = 10 - 1 = 9 L i t r e s [1 m^{3} = 10^{3} L]$
Implies that
$W = \frac{1 \times 9 \times 101}{1000} [1 L - a t o m = 101 J]$
Hence the answer is :
$W = 0.9 k J$

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