Question

A rigid container holds 1 kg of air initially at 4.8 bar and 150 degree Celsius the air is heated until its temperature is 200 degree Celsius determine the work done

Answer 1

Mass of air = 1 kg

Initial temperature T1 = 150 + 273 = 423 K

Initial pressure P1 = 4.8 bar

Final temperature T2 = 200 + 273 = 473 K

The container is rigid (volume is constant)

From the ideal gas equation

$P2=P1*T2/T1$

=4.8bar(473/423)

=5.37 bars

For isochoric process

Entropy change $S = Cv ln(\frac{T2}{T1})$

​= 0.08 kJ/kgK

Average temperature T = $\frac{(473 + 423)}{2} = 448 K$

Work done= $T \times S$

$= 448 K x 0.08 kJ/kgK = 35.9 kJ/kg$