A heat engine operates cyclically as described in the PV diagram below: Step 1.

Question

A heat engine operates cyclically as described in the PV diagram below: Step 1. Isothermal expansion at temperature TH and absorbs heat QH Step 2. Adiabatic expansion until temperature drops to Tc Step 3. Isothermal compression at Tc and exhausts heat lQcl Step 4. Adiabatic compression back to initial state 4 2 3 If the efficiency of the engine e=0.2, Tc=159K, the working gas has and there are n 58.0 moles of the gas, how much work does the gas do during step 2? Answer in multiples of 104 Joules

Explanation / Answer

work done in adiabatic process is (step2)

W = n*R/(1-gamma) (T2-T1)

W = R/(gamma-1) (T1-T2)

and efficiency is e = 1-T2/T1

given e = 0.2 , Tc = T2 = 159 k

so 0.2 = 1-159/T1

T1 = 198.75 k

substituting the values in the above formula  

W = 58*8.314/(1-5/3)(159-198.75) J

W = 28751.89 J

W = 2.8751*10^4 J

the workdone W = 2.8751*10^4 J

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