Question

Question #14:

An engine using 1 mol of an ideal gas initially at 18.5 L and 402 K performs a cycle
consisting of four steps:

1) an isothermal expansion at 402 K from 18.5 L to 41.2 L
2) cooling at constant volume to 273 K
3) an isothermal compression to its original volume of 18.5 L
4) heating at constant volume to its original temperature of 402 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 1

The efficiency of the engine through the four steps cycle is 83.4%.

What is the efficiency of a machine?

The efficiency of a machine is the ratio of useful work done by the machine and the work done on the machine expressed as a percentage.

The efficiency of the engine is given by:

$ɛ=\frac {W} {Q}$

where;

• W is the work done in the four steps,
• Q is the energy input

Since there at four steps in a cycle:

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

The work done in the first step (isothermal expansion) is given by:

$W_1=nRT_1\ln{\frac {V_2} { V_1}}$

where:

n= 1 mole, T1 = 402 K, V2 = 41.2 L, V1 = 18.5 L

Substituting the values:

$W_1 = 1 × 8.314 × 402 \times ln{\frac {41.2} {18.5}} = 2676.01 J$

Steps 2 and 4 are constant volume processes, therefore;

$W_2=W_4=0$

The work done in the third step (isothermal expansion) is given by:

$W_3=nRT_3\ln{\frac {V_4} { V_3}}$

where;

n = 1 mol, T3 = 273 K, V4 = 41.2 L, V3 = 18.5 L

Substituting the values:

$W_3 = 1 × 8.314 × 273 \times ln{\frac {41.2} {18.5}} = 1817.29 J$

$W_3 = 1 × 8.314 × 273 \times ln{\frac {41.2} {18.5}} = 1817.29 J$Heat enters the system only during steps (1) and (4).

The internal energy of the gas increases in step 4 but no work is done, while the internal energy is constant change in step 1 but work is done by the gas.

Thus;

$Q_2=Q_3=0$

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

$Q_1=W_1$

The heat that enters the system on step 4 at constant volume is given by:

$Q_4=C_V(T_4-T_3)$

where:

Cv =21 J/K, T3 = 273 K, T4 = 402 K

Substituting the values:

$Q_4= 21 × (402 - 273) = 2709 \: J$

Solving for efficiency, ɛ:

$ɛ=\frac {2676.01+ 0+0+1817.29} {2676.01+0+0+2709} = 83.4\%$

Therefore, the efficiency of the engine is 83.4%.

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