Problem: An insulated cylinder fitted with a piston contains 5 kg of air at 500 kPa and 1000 K. The air expands in an adiabatic process until its volume doubles. Calculate the work done by the air. Find: Work W done by air during expansion. Known: Mass of air m = 5 kg, initial pressure [latex]P_1[/latex] = 500 kPa, initial temperature [latex]T_1[/latex] = 1000 K, final volume [latex]ν_2 = 2ν_1[/latex] , adiabatic process so [latex]Q_{12}[/latex] = 0. Assumptions: Air is an ideal gas with constant specific heats, the process is reversible and adiabatic so ∆s = 0 Governing equations: Isentropic process (ideal gas, constant specific heats) [latex] \frac{T_2}{T_1}=\left\lgroup \frac{ν_1}{ν_2}\right\rgroup ^{(\gamma -1)}[/latex] First law [latex] ΔU=Q_{12}+W_{12}=Q_{12}+mc_ν(T_2-T_1) [/latex] Properties: Air at 1000 K has specific heat [latex]c_ν[/latex] = 0.855 kJ / kgK (Appendix 4), specific heat ratio of air at 1000 K γ = 1.336 (Appendix 4).

Solving for the final temperature, [latex]T_2=T_1\left\lgroup \frac{ν_1}{ν_2}\right\rgroup ^{(γ -1)}=1000 \ K \left\lgroup \frac{1}{2}\right\rgroup ^{(1.336-1)}=792.235 \ K. [/latex] For an adiabatic process [latex]Q_{12}[/latex] = 0, so the first law reduces to [latex]W_{12}= mc_ν(T_2-T_1)=5 \ kg \times 0.855 \ kJ/kgK \times (792.235 \ K -1000 \ K)=-888.195 \ kJ.[/latex] Answer: The gas does 888.2 kJ of work on the surroundings.

Question 6.5: Problem: An insulated cylinder fitted with a piston contains...

Energy, Entropy and Engines An Introduction to Thermodynamics

Problem: An insulated cylinder fitted with a piston contains 5 kg of air at 500 kPa and 1000 K. The air expands in an adiabatic process until its volume doubles. Calculate the work done by the air.

Find: Work W done by air during expansion.

Known: Mass of air m = 5 kg, initial pressure $P_1$ = 500 kPa, initial temperature $T_1$ = 1000 K, final volume $ν_2 = 2ν_1$ , adiabatic process so $Q_{12}$ = 0.

Assumptions: Air is an ideal gas with constant specific heats, the process is reversible and adiabatic so ∆s = 0

Governing equations:
Isentropic process (ideal gas, constant specific heats) $\frac{T_2}{T_1}=\left\lgroup \frac{ν_1}{ν_2}\right\rgroup ^{(\gamma -1)}$

First law $ΔU=Q_{12}+W_{12}=Q_{12}+mc_ν(T_2-T_1)$

Properties: Air at 1000 K has specific heat $c_ν$ = 0.855 kJ / kgK (Appendix 4), specific heat ratio of air at 1000 K γ = 1.336 (Appendix 4).

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Solving for the final temperature,

$T_2=T_1\left\lgroup \frac{ν_1}{ν_2}\right\rgroup ^{(γ -1)}=1000 \ K \left\lgroup \frac{1}{2}\right\rgroup ^{(1.336-1)}=792.235 \ K.$