This problem is an expansion of Example 2-3. The table below lists 10 sets of conditions—5 temperatures at a constant P, and five pressures at a constant T. For each T and P, find: The specific volume of steam, from the steam tables The specific volume of steam, from the ideal gas law Comment on

Question

This problem is an expansion of Example 2-3. The table below lists 10 sets of conditions—5 temperatures at a constant P, and five pressures at a constant T. For each T and P, find:

The specific volume of steam, from the steam tables
The specific volume of steam, from the ideal gas law

Comment on the results. Under what circumstances does departure from ideal gas behavior increase?

[latex]\begin{gathered}\begin{matrix} \hline ext{Temperature}& ext{Pressure}\\hline200^{\circ}C & 5 m\, bar\\hline 300^{\circ}C & 5 m\, bar\\hline 400^{\circ}C & 5 m\, bar\\hline 500^{\circ}C & 5 m\, bar\\hline 1000^{\circ}C & 5 m\, bar\\hline 250^{\circ}C & 0.1 m\, bar\\hline 250^{\circ}C & 1 m\, bar\\hline 250^{\circ}C & 5 m\, bar\\hline 250^{\circ}C & 10 m\, bar\\hline 250^{\circ}C & 25 m\, bar\\hline \end{matrix} \end{gathered}[/latex]

Accepted Answer

Here we will display the calculations and reasoning behind the first set of conditions (bolded values):

		Specific Volume ([latex]H_2O[/latex])
Temperature	Pressure	Steam Tables (m³/kg)	Ideal Gas Law (m³/kg)	% Difference
200°C	5 bar	0.425	0.437	2.82
300°C	5 bar	0.523	0.529	1.15
400°C	5 bar	0.617	0.621	0.65
500°C	5 bar	0.711	0.713	0.28
1000°C	5 bar	1.175	1.175	0.00
250°C	0.1 bar	24.136	24.132	-0.02
250°C	1 bar	2.406	2.413	0.29
250°C	5 bar	9.474	0.483	1.90
250°C	10 bar	0.233	0.241	3.43
250°C	25 bar	0.087	0.097	11.49

(From Steam tables) Steam at 200°C and 5 bar

(From Ideal Gas Law)

[latex]\begin{aligned}& \mathrm{PV}=\mathrm{nRT} ightarrow \underline{\mathrm{V}}=\frac{\mathrm{RT}}{\mathrm{P}} \& \underline{\mathrm{V}}=\frac{\left(0.08206 \frac{\mathrm{L\,atm}}{\mathrm{mol} \,\mathrm{K}} ight)(473.15 \mathrm{~K})}{5 \mathrm{bar}}\left(\frac{1.01325 \,\mathrm{bar}}{1 \mathrm{~atm}} ight)\left(\frac{1 \mathrm{~mol}}{18.02 \mathrm{~g}} ight)\left(\frac{1 \mathrm{~m}^{3}}{1000 \mathrm{~L}} ight)\left(\frac{1000 \mathrm{~g}}{1 \mathrm{~kg}} ight) \&=\mathbf{0 . 4 3 7 \frac { \mathbf { m } ^ { 3 } } { \mathbf{ kg } }}\end{aligned}[/latex]

Departures from ideal gas behavior increase with increasing pressure and with decreasing temperature, since both of these lead to lower volume, and lower volume means more significant intermolecular interaction

Question 2.14: This problem is an expansion of Example 2-3. The table below...

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