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Example 9.8: Volume of a Gas Sample

The sample of gas in Figure 13 has a volume of 15.0 mL at a pressure of 13.0 psi. Determine the pressure of the gas at a volume of 7.5 mL, using:
(a) the \(\ce{P-V}\) graph in Figure 13
(b) the 1/P vs. V graph in Figure 13
(c) the Boyle’s law equation
Comment on the likely accuracy of each method.

Solution

\(\mathrm{psi}\) \(= 6894.76\ \mathrm{Pa}\)


\(V_{\mathrm{1}}\) \(= 15.0\ \mathrm{mL}\)


\(P_{\mathrm{1}}\) \(= 13.0 \cdot \mathrm{psi}\)

\(\ \ \ =13.0 \cdot 6894.76\ \mathrm{Pa}\)

\(\ \ \ =8.96\times 10^{4}\ \mathrm{Pa}\)


\(V_{\mathrm{2}}\) \(= 7.5\ \mathrm{mL}\)


\(P_{\mathrm{2}}\) = ?


(a) Estimating from the \(\ce{P-V}\) graph gives a value for P somewhere around 27 psi.
(b) Estimating from the 1/P versus V graph give a value of about 26 psi.
(c) From Boyle’s law, we know that the product of pressure and volume \(\ce{(PV)}\) for a given sample of gas at a constant temperature is always equal to the same value. Therefore we have


Solving for the unknown P2:

\(P_{\mathrm{2}}\) \(= P_{\mathrm{1}} \cdot \dfrac{V_{\mathrm{1}}}{V_{\mathrm{2}}}\)

\(\ \ \ =8.96\times 10^{4}\ \mathrm{Pa} \cdot \dfrac{15.0\ \mathrm{mL}}{7.5\ \mathrm{mL}}\)

\(\ \ \ =8.96\times 10^{4}\ \mathrm{Pa} \cdot 2.000\)

\(\ \ \ =1.79\times 10^{5}\ \mathrm{Pa}\)


\(P_{\mathrm{2}}\) \(= \dfrac{P_{\mathrm{2}}}{\mathrm{psi}}\mathrm{\ \mathrm{psi}}\)

\(\ \ \ =\dfrac{1.79\times 10^{5}\ \mathrm{Pa}}{6894.76\ \mathrm{Pa}}\mathrm{\ \mathrm{psi}}\)

\(\ \ \ =\dfrac{1.79\times 10^{5}\ \mathrm{Pa}}{6894.760\ \mathrm{Pa}}\mathrm{\ \mathrm{psi}}\)

\(\ \ \ =26.0\mathrm{\ \mathrm{psi}}\)


It was more difficult to estimate well from the \(\ce{P-V}\) graph, so (a) is likely more inaccurate than (b) or (c). The calculation will be as accurate as the equation and measurements allow.