Chapter 2, Problem 2.91E

The Dieterici equation of state for one mole of gas is

$p=RT \frac{e^{\frac{-a}{VRT}}}{\bar{V}-b}$

Where a and b are constants determined experimentally. For NH₃(g), a = 10.91 atm. L² and b = 0.0401 L. Plot the pressure of the gas as the volume of 1.00 mol of NH₃(g) expands from 22.4 L to 50.0 L at 273 K, and numerically determine the work done by the gas by measuring the area under the curve.

Interpretation Introduction

Interpretation:

The Dieterici equation of state for one mole of gas is

$p=RT \frac{e^{\frac{-a}{VRT}}}{\overline{V}-b}$

Where a and b are constants determined experimentally. For NH₃(g), a = 10.91 atm. L² and b = 0.0401 L. The pressure of the gas as the volume of 1.00 mol of NH₃(g) expands from 22.4 L to 50.0 L at 273 K is to be plotted and numerically the work done by the gas by measuring the area under the curve is to be determined.

Concept introduction:

The ideal gas law considered the molecules of a gas as point particles with perfectly elastic collisions among them in nature. This works importantly well for gases at dilution and at low pressure in many experimental calculations. But the gas molecules are not performing as point masses, and there are situations where the properties of the gas molecules have measurable effect by experiments. Thus, a modification of the ideal gas equation was coined by Johannes D. van der Waals in 1873 to consider size of molecules and the interaction forces among them. Berthelot modified the van der Waals equation as modified Berthelot model of state and further changes was made, and the equation was provided as Dieterici equation of state. The significant advantages of this equation, such as a more realistic critical compressibility factor are documented.

Answer to Problem 2.91E

The Dieterici equation of state for one mole of gas is

$p=RT \frac{e^{\frac{-a}{VRT}}}{\overline{V}-b}$ using the equation pressure of one mole of gas at 22.4 L is calculated as 0.9794 atm and pressure of one mole of gas at 50 L is calculated as 0.4435 atm. Moreover, the work done by the ammonia gas is calculated as 1498 J.

Explanation of Solution

The Dieterici equation of state for one mole of gas is $p=RT \frac{e^{\frac{-a}{VRT}}}{\overline{V}-b}$

Given,

a = 10.91 atm. L²

b = 0.0401 L

volume of gas initial = 22.4 L

volume of gas final = 50.0 L

temperature of system = 273 K

pressure at 22.4 L is calculated as,

$\frac{a}{VRT}= \frac{10.91 atm L^2}{22.4 L x 0.0820 \frac{L atm}{K mol} x 273 K}$

$=0.0217 mol$

$e^{\frac{-a}{VRT}}= e^{-\left(0.0217 mol\right)}$

$=0.9785 mol$

$p=\frac{0.0820 \frac{L. atm}{K. mol} x 273 K x 0.9785 mol}{\left(22.41-0.0401\right) L}$

∴ pressure at 22.4 L = 0.9794 atm

Similarly, pressure at 50 L is calculated as follows;

$\frac{a}{VRT}= \frac{10.91 atm L^2}{50 L x 0.0820 \frac{L atm}{K mol} x 273 K}$

$=0.00974 mol$

$e^{\frac{-a}{VRT}}= e^{-\left(0.00974 mol\right)}$

$=0.9903 mol$

$p=\frac{0.0820 \frac{L. atm}{K. mol} x 273 K x 0.9903 mol}{\left(50-0.0401\right) L}$

∴ pressure at 50 L = 0.4435 atm

From the graph the pressure difference can be calculated as;

0.5359 atm.

the work done by the gas by measuring the area under the curve is determined as;

$\begin{array}{l}{\text{w = P}}_{\text{ex}}\text{ΔV}\hfill \\ \text{ = 0}\text{.5359 atm x 27}\text{.6 L}\hfill \\ \text{ = 14}\text{.79 L}\text{. atm}\hfill \\ \text{w = 1498 J}\hfill \end{array}$

Conclusion

Thus, the pressure of the gas is plotted against volume and the work done in expansion is calculated.