A piston–cylinder device contains 0.78 kg of nitrogen gas at 140 kPa and 37°C. The gas is now compressed slowly in a polytropic process during which PV1.3 = constant. The process ends when the volume is reduced by one-half. Determine the entropy change of nitrogen during this process. The gas constant of nitrogen is R = 0.2968 kJ/kg·K. The constant volume specific heat of nitrogen at room temperature is cv = 0.743 kJ/kg·K. (Round the final answer to five decimal places.)

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.

visit this site

explanation:

oil flow is a highly convective flow compared to water flow at normal temperature and pressure.