Helium contained in a cylinder at ambient conditions, 100 kPa, 20°C, is compressed in a reversible isothermal process to 600 kPa, after which the gas is expanded back to 100 kPa in a reversible adiabatic process. Calculate the net work per kilogram of helium.

Helium contained in a cylinder at ambient conditions, 100 kPa, 20°C, is compressed in a reversible isothermal process to 600 kPa, after which the gas is expanded back to 100 kPa in a reversible adiabatic process. Calculate the net work per kilogram of helium. Correct Answer -623.6 kJ/kg

The adiabatic reversible expansion gives constant s T3 = T2(P3/P2)^ = 293.15 (100/600)^0.4 = 143.15 K The isothermal process: 1w2 = -RT1 ln(P2/P1) = -2.0771 × 293.15 × ln(600/100) = -1091.0 kJ/kg The adiabatic process: 2w3 = CVo(T2-T3) = 3.116 (293.15 – 143.15) = +467.4 kJ/kg The net work is the sum: w(NET) = -1091.0 + 467.4 = -623.6 kJ/kg.

Related Questions

On a P-V diagram of an ideal gas, suppose a reversible adiabatic line intersects a reversible isothermal line at point A. Then at a point A, the slope of the reversible adiabatic line $${\left( {\frac{{\partial {\text{P}}}}{{\partial {\text{V}}}}} \right)_{\text{S}}}$$  and the slope of the reversible isothermal line $${\left( {\frac{{\partial {\text{P}}}}{{\partial {\text{V}}}}} \right)_{\text{T}}}$$  are related as (where, $${\text{y}} = \frac{{{{\text{C}}_{\text{p}}}}}{{{{\text{C}}_{\text{v}}}}}$$  )