(a) Determine the specific volume of nitrogen gas at 8.5 MPa and 139 K based on the generalized compressibility chart and compare this result with the experimental value of 0.002002 m³/kg to (b).
(b) Determine the error involved.
(R = 0.2968 kPa. m³/kg. K, Tcr=126.2 K, Pcr=3.39 MPa)

V = 0.002524m³/kg

%Error = 0.2851

Explanation:

Using the ideal gas equation, we can determine the specific volume of nitrogen as

PV = RT

V = RT/P

Given the following data:

R = 296.8j/kg.k

T = 139k

P = 8.5Mpa

Therefore, V = (296.8j/kg.k)×(139k)/8500000

V = 0.004853m³/kg

Using the compressibility chart, we can see that the compressibility factor of Nitrogen at this state is equal to 0.52. Using this, we can find out the specific volume of nitrogen to be

Z = PV/RT

0.52 = (8500000Pa)V/(296.8J/kg⋅K)(139K)

V = 0.52×296.87×139/8500000

V = 0.002524m³/kg

%Error = V(ideal)-V(actual)/V(actual)

%Error = 0.002524-0.002002/0.002002

%Error = 0.2609

Error for this compared with the actual value is

%Error = V(ideal)-V(actual)/V(actual)

%Error = (0.004853-0.002002)/0.004853

%Error = 0.2851

the clapeyron equation (also called the clausius-clapeyron equation) relates the slope of a reaction line on a phase diagram to fundamental thermodynamic properties.

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