Answer to Question #87604 in Molecular Physics | Thermodynamics for Lesly

Question

An engine using 1 mol of an ideal gas initially at 19 L and 331 K performs a cycle consisting of four steps:
1) an isothermal expansion at 331 K from 19 L to 39.4 L ;
2) cooling at constant volume to 218 K ;
3) an isothermal compression to its original volume of 19 L; and
4) heating at constant volume to its original temperature of 331 K .
Find its efficiency. Assume that the heat capacity is 21 J/K and the universal gas constant is 0.08206 L · atm/mol/K = 8.314 J/mol/K.
Answer in units of %.

Answer 1

The efficiency is given by

$$ɛ=\frac {W} {Q} (1)$$

where W is the work done during four steps, Q is the energy that was passed

In this case, we can write

$$ɛ=\frac {W_1+ W_2+W_3+W_4} {Q_1+Q_2+Q_3+Q_4} (2)$$

The work done during the isothermal expansion (1) is given by:

$$W_1=nRT_1\ln{\frac {V_2} { V_1}} (3)$$

where n=1 mol, T₁=331 K, V₂=39.4 L, V₁=19 L

Steps 2 and 4 are constant volume processes

$$W_2=W_4=0 (4)$$

The work done during the isothermal expansion (3) is given by:

$$W_4=nRT_3\ln{\frac {V_4} { V_3}} (5)$$

where n=1 mol, T₃=218 K, V₄=39.4 L, V₃=19 L

Heat enters the system only during processes (1) and (4) because the internal energy of the gas increases in step 4 while no work is done, and because the internal energy does not change during step 1 while work is done by the gas:

$$Q_2=Q_3=0 (6)$$

The heat that enters the system during this isothermal expansion is given by:

$$Q_1=W_1 (7)$$

The heat that enters the system during the constant volume step 4 is given by

$$Q_4=C_V(T_3-T_4) (8)$$

where C_v=21 J/K, T3=218 K, T4=331 K

We put (3)- (8) in (2) and got:

ɛ=15.2%

Answer:

15.2%