Complete Question:
A heat engine with 0.300 mol of a monatomic ideal gas initially fills a 1000 cm3 cylinder at 500 K . The gas goes through the following closed cycle: - Isothermal expansion to 5000 cm3. - Isochoric cooling to 400 K . - Isothermal compression to 1000 cm3. - Isochoric heating to 500 K .
a) what is the work for one cycle
b) what is the thermal efficiency
Answer:
a) Work done for 1 cycle = 402.13
b) Thermal efficiency = 20%
Explanation:
Number of moles, n = 0.300 mol
Initial Volume, V₁ = 1000 cm³
Temperature, T = 500 K
Isothermal expansion to 5000 cm³
Final volume, V₂ = 5000 cm³
R = 8.314 J/ mol.K
Work done, W = nRT ln(V₂/V₁)
W = (0.3 * 8.314 * 500) * ln(5000/1000)
W = 1247.1 * ln5
W₁ = 2007.13 J
Isochoric cooling
In an Isochoric process, volume is constant i.e. V₂ = V₁ = V
W = nRT ln(V/V)
But ln(V/V) = ln 1 = 0
Work done, W₂ = 0 Joules
Isothermal Compression to 1000 cm³
V₂ = 1000 cm³
V₁ = 5000 cm³
W = nRT ln(V₂/V₁)
W = 0.3 * 8.314 * 400 ln(1000/5000)
W₃ = -1605 J
Isochoric heating to 500 K
Since there is no change in volume, no work is done
W₄ = 0 J
a) Work done for 1 cycle
W = W₁ + W₂ + W₃ + W₄
W = 2007.13 + 0 + 0 -1605+0
W = 402.13 Joules
b) Thermal efficiency
Thermal efficiency = (Net workdone for 1 cycle)/(Heat absorbed)
Heat absorbed = Work done due to thermal expansion = 2007.13 J
Thermal efficiency = 402.13/2007.13
Thermal efficiency = 0.2
Thermal efficiency = 0.2 * 100% = 20 %
