Question

A 2 m3 insulated rigid tank contains 3 kg of nitrogen at 90 kPa. Now work is done on the system until the pressure in the tank rises to 175 kPa. Determine the entropy change of nitrogen in kJ/K during this process assuming constant specific heats.

Answer 1

Answer: $\Delta S$ = 1.47kJ/K

Explanation: Entropy is the measure of a system's molecular disorder, i.e, the unuseful work a system does.

The nitrogen gas in the insulated tank can be described as an ideal gas, so it can be used the related formulas.

For the entropy, the ratio of initial and final temperatures is needed and as volume is constant, we use:

$\frac{P_{1}}{T_{1}} =\frac{P_{2}}{T_{2}}$

$\frac{P_{2}}{P_{1}} =\frac{T_{2}}{T_{1}}$

$\frac{T_{2}}{T_{1}} =\frac{175}{90}$

$\frac{T_{2}}{T_{1}} =1.94$

Specific Heat is the quantity of heat required to increase the temperature 1 degree of a unit mass of a substance. Specific heat of nitrogen at constant volume is $c_{v}=$ 0.743kJ/kg.K

The change in entropy is calculated by

$\Delta S= m[c_{v}ln(\frac{T_{2}}{T_{1}})-Rln(\frac{V_{2}}{V_{1}} )]$

For the nitrogen insulated in a rigid tank:

$\Delta S= m[c_{v}ln(\frac{T_{2}}{T_{1}})]$

Substituing:

$\Delta S= 3[0.743ln(1.94)]$

$\Delta S=$ 1.47

The entropy change of nitrogen in an insulated rigid tank is 1.47kJ/K