A fixed amount of ideal gas changes from P₁ = 1.45 × 10^5 Pa, V₁ = 0.0029 m³, T₁ = 308 K to P₂ = 1.76 × 10^5 Pa and T₂ = 357 K. Determine V₂.

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0.00277 m³ Start from PV = nRT. The amount of gas and R do not change, so PV/T has the same value in both states; the supplied absolute temperatures are already in kelvin. Equating the states gives P₁V₁/T₁ = P₂V₂/T₂, then rearranging isolates V₂ = P₁V₁T₂/(P₂T₁). Substitution gives V₂ = [(1.45 × 10^5)(0.0029)(357)]/[(1.76 × 10^5)(308)] = 0.00277 m³. As a check, V₂/V₁ = (P₁/P₂)(T₂/T₁) = 0.955; this combines the volume-reducing pressure rise with the volume-increasing temperature rise.

Fixed-gas state change 9

A fixed amount of ideal gas changes from P₁ = 1.45 × 10⁵ Pa, V₁ = 0.0029 m³, T₁ = 308 K to P₂ = 1.76 × 10⁵ Pa and T₂ = 357 K. Determine V₂.

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