Question 7.13: Problem: Calculate the pressure of nitrogen that has a speci...
Problem: Calculate the pressure of nitrogen that has a specific volume ν = 0.025 m³ / kg and a temperature of 252 K using (a) the ideal gas equation and (b) the van der Waals equation of state.
Find: Pressure of nitrogen P.
Known: Specific volume ν = 0.025 m³ / kg, temperature T = 252 K.
Governing Equations:
Ideal gas equation P\bar{ν}=R_uT
van der Waals equation \left\lgroup P+\frac{a}{\bar{\nu }^2}\right\rgroup (\bar{\nu}-b)=R_uT
Properties: For nitrogen, the critical temperature T_C = 126.2 K (Table 7.1), critical pressure P_C = 3.39 MPa (Table 7.1) and molar mass M = 28.013 kg / kmol (Appendix 1), van der Waals constants a = 137 kPa m^6\ / kmol^2 and b = 0.0386 m³ / kmol (Table 7.5).
| Table 7.1 Critical temperature (T_c), pressure (P_c) and volume (v_c) of common substances. | ||||
| Substance | Formula | Temperature, T_c (K) | Pressure, P_c (MPa) | Volume, v̄_c (m^3 /kmol) |
| Air | — | 132.5 | 3.77 | 0.0883 |
| Ammonia | NH_3 | 405.5 | 11.28 | 0.0724 |
| Argon | Ar | 151 | 4.86 | 0.0749 |
| Carbon dioxide | CO_2 | 304.2 | 7.39 | 0.0943 |
| Helium | He | 5.3 | 0.23 | 0.0578 |
| Methane | CH_4 | 191.1 | 4.64 | 0.0993 |
| Nitrogen | N_2 | 126.2 | 3.39 | 0.0899 |
| Oxygen | O_2 | 154.8 | 5.08 | 0.078 |
| Water | H_2O | 647.3 | 22.09 | 0.0568 |
| Table 7.5 Values of van der Waals constants for common gases. | |||
| Substance | Formula | a (kPa m^6 / kmol^2 ) | b (m^3 /kmol) |
| Air | — | 136 | 0.0365 |
| Carbon dioxide | CO_2 | 364.7 | 0.0428 |
| Nitrogen | N_2 | 137 | 0.0386 |
| Hydrogen | H_2 | 24.5 | 0.0263 |
| Water | H_2O | 554 | 0.0305 |
| Oxygen | O_2 | 136.9 | 0.0319 |
| Helium | He | 3.47 | 0.0238 |
| Argon | Ar | 135 | 0.0320 |
| Methane | CH_4 | 229.3 | 0.048 |
The molar specific volume of nitrogen is
\bar{\nu}=\nu M=0.025 \ m^3/kg \times 28.013 \ kg/kmol=0.700 \ 33 \ m^3/kmol.
(a) Ideal gas equation ,
P=\frac{R_uT}{\bar{\nu}}=\frac{8.314 \ kJ /kmolK \times 252 \ K}{0.700 \ 33 \ m^3/kmol}=2991.6 \ kPa = 2.99 \ MPa.
(b) van der Waals equation,
P=\frac{R_uT}{(\bar{\nu}-b)}-\left\lgroup \frac{a}{\bar{\nu}^2} \right\rgroup , \\ P= \frac{8.314 \times 10^3 J/kmolK \times 252 \ K}{(0.700 \ 33 \ m^3/kmol – 0.0386 \ m^3/kmol)}-\left\lgroup \frac{137 \times 10^3 Pam^6/kmol^2}{(0.700 \ 33 \ m^3/kmol)^2}\right\rgroup \ , \\ P=2.8868 \ MPa.
Answer: The pressure calculated using the (a) ideal gas equation is 2.99 MPa, (b) van der Waals equation is 2.89 MPa.