Question

On the basis of the ideal gas equation and van der Waals equation, the temperatures of a gas at pressure 10 MPa and specific volume 0.005 m$^3$/kg would be, respectively
(Assume gas constant $R = 0.3$ kJ/(kg K), $a = 0.18$ m$^6$ kPa/kg$^2$ and $b = 0.0014$ m$^3$/kg)

• A. 166.67 K and 235.89 K
• B. 166.67 K and 206.40 K
• C. 166.67 K and 267.21 K
• D. 166.67 K and 240.90 K

Hint:

Ideal gas law underestimates temperature at high pressures; Van der Waals corrections account for molecular attraction and finite volume.

Answer 1

The Correct Option is B

Using the ideal gas law:
\[ T_{\text{ideal}} = \frac{Pv}{R} = \frac{(10{,}000\ \text{kPa})(0.005)}{0.3} = 166.67\ \text{K}. \] For the Van der Waals gas:
\[ \left(P + \frac{a}{v^2}\right)(v - b) = R T. \] Compute the correction terms:
\[ \frac{a}{v^2} = \frac{0.18}{(0.005)^2} = 7200\ \text{kPa}, \] \[ P + \frac{a}{v^2} = 10000 + 7200 = 17200\ \text{kPa}, \] \[ v - b = 0.005 - 0.0014 = 0.0036. \] Thus, \[ T_{\text{vdW}} = \frac{17200 \times 0.0036}{0.3} = 206.40\ \text{K}. \]