Question

An engine using 1 mol of an ideal gas ini- tially at 21.7 L and 326 K performs a cycle consisting of four steps:
1) an isothermal expansion at 326 K from 21.7 L to 42.1 L;
2) cooling at constant volume to 186 K ;
3) an isothermal compression to its original volume of 21.7 L; and
4) heating at constant volume to its original
temperature of 326 K .
Find its efficiency. Assume that the
heat capacity is 21 J/K and the univer- sal gas constant is 0.08206 L · atm/mol/K = 8.314 J/mol/K.
Answer in units of %.

Answer 1

Answer to Question #88246,Physics / Molecular Physics | Thermodynamics

An engine using 1 mol of an ideal gas initially at 21.7 L and 326 K performs a cycle consisting of four steps:

1) an isothermal expansion at 326 K from 21.7L to 42.1 L ;

2) cooling at constant volume to 186 K ;

3) an isothermal compression to its original volume of 21.7L; and

4) heating at constant volume to its original temperature of 326K .

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 %.

Solution:n=1, V₁=21.7L

i) At T₁=326K, Volume changes from V₁=21.7L to V₂=42.1L (Say the path A to B)

It is a constant temperature process where the gases are expanded from volume Vₐ to V₈. Thus is a isothermal expansion. In this case heat is absorbed by the gas.

Thus work is done by the gas when heat is added.

ii) Cooling at constant volume (Say path B to C). Temperature decreases from T₁=326K to T₂=186K.

In this case no any heat is added and since it is constant volume process, the work is done is zero

iii) At temperature T₂=186K, the volume is decreases from Vc=42.1L to Vd=21.70L (say path C to D. It is a constant temperature process where the gas is compressed from V₂ to V₁.

Since the volume is decreases so work is done on the gas but no any heat is added.

iv) Heating at constant volume to its original temperature of 326 K (Say path D to A). Volume is constant, but temperature is increases from T₂=186K to T₁=326K. Work is done on the gas and heat is rejected.

Thus efficiency of the engine= Total work done/Net heat added

Path AB,

Work done=nRThotln(VB/VA)

Path BC,

Work done=0

Path CD,

Work done=nRTcoldln(VD/VC)

Path DA,

Work done =0

Heat absorbed,

Path AB,

$Q_{1} = W1 = nRT_{hot}\ln (V_{B} / V_{A})$

Path DA,

$Q_{4} = C_{v}$ deltaT=Cv(Thot-Tcold)

Efficiency=

\left\{ \frac{nRT_{hot} \ln \left( \frac{V_B}{V_A} \right) + nRT_{cold} \ln \left( \frac{V_D}{V_C} \right)}{\left\{ \frac{nRT_{hot} \ln \left( \frac{V_B}{V_A} \right) + C_v (T_{hot} - T_{cold})}{nR} \right\} \right\} \left\{ \frac{T_{hot} \ln \left( \frac{V_B}{V_A} \right) + T_{cold} \ln \left( \frac{V_D}{V_C} \right)}{\left\{ \frac{T_{hot} \ln \left( \frac{V_B}{V_A} \right) + C_v (T_{hot} - T_{cold})}{nR} \right\} \right\} \left\{ \frac{326 \ln \left( \frac{42.1}{21.7} \right) + 186 \ln \left( \frac{21.7}{42.1} \right)}{\left\{ \frac{326 \ln \left( \frac{42.1}{21.7} \right) + 21 \left( \frac{326 - 186}{1 \times 8.314} \right)}{1 \times 8.314} \right\} \right\}

= {326×0.66 - 186×0.66} / {326×0.66 + 21×16}

=92.4/551.16

=16.76%

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