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Answer: about 152 J

For an ideal gas \(P = \frac{2}{3}\dfrac{E}{V}\), so \(E = \frac{3}{2}PV\). At S.T.P., \(P = 1.013 \times 10^5\) N/m\(^2\) and \(V = 1\) litre \(= 10^{-3}\) m\(^3\):

\[E = \frac{3}{2} \times 1.013 \times 10^5 \times 10^{-3} = 151.95 \text{ J} \approx 152 \text{ J}\]
Doubtly AI · 28 Oct 2026 AI-verified · Oct 2026 Report
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