Q.324. Thermodynatics - First Law, Work done, Heat absorbed.

Question 324.
1) A reversible Carnot cycle heat engine operates between an upper temperature T₁ and a lower temperature T₂. It consists of four steps that take the engine through the cycle: (T₁, V₁)→ (T₁,V₁ʹ)→(T₂, V₂)→(T₂, V₂ʹ)→ (T₁, V₁). The volumes in the cycle satisfy the relation: ( V₂ʹ - b)/ (V₂- b)= (V₁- b)/(V₁ʹ - b). Compute the work w for the entire cycle.
 2) The working substance of the engine in part 1 is one mole of a gas with equation of state, P(V-b)= RT, where b is a constant. Compute the heat absorbed by the system in the process (T₁, V₁)→ (T₁,V₁ʹ).

Answer 324.
Carnot cycle is made up of four reversible processes of an ideal gas

i) isothermal expansion,
ii) adiabatic expansion,
iii) isothermal compression and
iv) adiabatic compression.

1)
Net work done in adiabatic processes is zero.
Net work done in isothermal processes
= W₁ + W₂
= RT₁ ln[(V₁' - b)/(V₁ - b)] + RT₂ ln[(V₂' - b)/(V₂ - b)]
= RT₁ ln[(V₁' - b)/(V₁ - b)] - RT₂ ln[(V₁' - b)/(V₁ - b)]
= R(T₁ - T₂) ln[V₁' - b)/(V₁ - b)]
which is the net work done for the entire cycle.

2)
Heat absorbed in the process (T ₁, V₁)→ (T ₁,V₁ʹ).
Q
= nCvdT + W₁
= 0 + RT₁ln[V₁' - b)/(V₁ - b)].

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