Answer:
The entropy change of nitrogen during this process. is - 0.32628 kJ/K.
Explanation:
Solution
Given that:
A piston cylinder device contains =0.78 kg of nitrogen gas
Temperature = 37°C
The nitrogen gas constant of R = 0.2968 kJ/kg.K
At room temperature cv = 0.743 kJ/kg.K
Now,
We assume that at specific condition the nitrogen can be treated as an ideal gas
Nitrogen has a constant volume specific heat at room temperature.
Thus,
From the polytropic relation, we have the following below:
T₂/T₁ =(V₁/V₂)^ n-1 which is,
T₂ = T₁ ((V₁/V₂)^ n-1
= (310 K) (2)^1.3-1 = 381.7 K
So,
The entropy change of nitrogen is computed as follows:
ΔSN₂ = m (cv₁ avg ln T₂/T₁ + ln V₂/V₁)
= (0.78) ((0.743 kJ/kg .K) ln 381.7 K/310K + (0.2968 kJ/kg. K) ln (0.5))
= 0.57954 * 0.2080 + (-0.2057)
= 0.12058 + (-0.2057) = -0.32628
Therefore the entropy change of nitrogen during this process. is - 0.32628 kJ/K.
