Name: Sign Conventions and Work Done in Thermodynamics
Uploaded: 2019-07-19
Duration: 6 min 11 s
Description: In the field of thermodynamics, either heat is given to a system or by the system. Similarly, work may be done by the system or on the system. Let's learn more!

Question

Prove that the efficiency of a heat engine based on a cycle consisting of two isotherms and two isochors is lower than the efficiency of Carnot's heat engine operating with the same heater and cooler.

Accepted Answer

Let us analyse the operation of the heat engine based on the cycle formed by two isotherms and two isochors (Fig.). Suppose that the temperature of the cooler (corresponding to the lower isotherm) is $T_{1}$ , and the temperature of the heater (corresponding to the upper isotherm) is $T_{2}$ . On the isochoric segment $1 - 2$ , the gas volume does not change, i.e. no work is done, but the temperature increases from $T_{1}$ to $T_{2}$ . It means that a certain amount of heat $Q_{1}$ is supplied to the gas. On the isothermal segment $2 - 3$ , the internal energy of

the gas remains constant, and the entire amount of heat $Q_{2}$ supplied to the gas is spent on doing work: $Q_{2} = W_{2}$ .

On the isochoric segment $3 - 4$ , the gas temperature returns to its initial value $T_{1}$ , i.e. the anount of heat $U_{1}$ is removed from the gas. On the isothermal segment $4 - 1$ , the work done by the gas is negative, which means that some amount of heat is taken away from the gas. Thus, the total amount of heat supplied to the gas per cycle is $Q_{1} + W_{2}$ . Figure shows that the work done by the gas per cycle is the sum of the positive work $W_{2}$ on the segment $2 - 3$ and the negative work $W_{4}$ on the segment $4 - 1$ .

Let us compare the pressures at the points corresponding to equal volumes on the segments $4 - 1$ and $2 - 3$ . The Gay-Lussac law indicates that the ratio of these pressures is $\frac{T_{1}}{T_{2}}$ , and hence the work done by the gas is $W_{4} = - (\frac{T_{1}}{T_{2}}) W_{2}$ . The total work per cycle is given by

$W = W_{2} + W_{4} = (1 - \frac{T_{1}}{T_{2}}) W_{2},$
and the efficiency is

$η = \frac{W}{Q_{1} + W_{2}} = \frac{1 - \frac{T_{1}}{T_{2}}}{1 + \frac{Q_{1}}{W_{2}}} < 1 - \frac{T_{1}}{T_{2}} .$

Therefore, the efficiency of the heat engine based on the cycle consisting of two isotherms and two isochors is lower than the efficiency $η_{c} = 1 - \frac{T_{1}}{T_{2}}$ of Carnot's heat engine.

Prove that the efficiency of a heat engine based on a cycle consisting of two isotherms and two isochors is lower than the efficiency of Carnot's heat engine operating with the same heater and cooler.

Important Questions on Heat and Molecular Physics

Learn and Prepare for any exam you want !