Question

A reversible engine contains 0.350 mol of ideal monatomic gas, initially at 586 K and confined to a volume of 2.42 L . The gas undergoes the following cycle:
⋅ Isothermal expansion to 4.74 L
⋅ Constant-volume cooling to 252 K
⋅ Isothermal compression to 2.42 L
⋅ Constant-volume heating back to 586 K

Determine the engine's efficiency in percents, defined as the ratio of the work done to the heat absorbed during the cycle.

Answer 1

For a monatomic ideal gas, C_v = 1.5 R; T₁ = 586 K; T₂= 252 K

1. Isothermal expansion to 4.74 L :

For an isothermal change, dE = 0

dE = q + w = q-pdV            [as, w = -pdV]

q = pdV = nRT₁ ln(V₂/V₁) = nRT₁ ln (4.74/2.42)                           (q is positive , so heat is absorbed)

2. Constant-volume cooling to 252 K :

Here, dV = 0, So w = 0

dE = q = nC_v dT =nC_v (252-586) = - 334nC_v                (q is negative, heat is released)

3. Isothermal compression to 2.42 L :

For an isothermal change, dE = 0

dE = q + w = q-pdV            [as, w = -pdV]

q = pdV = nRT₂ ln (2.42/4.74)                       (q is negative, heat is released)

4. Constant-volume heating back to 586 K

Here, dV = 0, So w = 0

dE = q = nC_v dT =nC_v (586-252) = 334nC_v                  (q is positive , so heat is absorbed)

Total work done

= nRT₁ ln (4.74/2.42) + nRT₂ ln (2.42/4.74)

= 653.38 J            [put the value of n, R and T]

Total heat absorbed

= = nRT₁ ln (4.74/2.42) + 334nC_v

= 2604.2 J

Efficiency

= [(Total work done)/( Total heat absorbed)] *100

= (653.38/2604.2) * 100

= 25%