Example 9.6: Predicting Change in Volume with Temperature
A sample of carbon dioxide,
\(\ce{CO2}\), occupies
0.300 L at
10 °C and
750 torr. What volume will the gas have at
30 °C and
750 torr?
Solution
\(V_{\mathrm{1}}\) \(= 0.300\ \mathrm{L}\)
\(T_{\mathrm{1}}\) \(= 10.\ \mathrm{°aC}\)
\(T_{\mathrm{2}}\) \(= 30.\ \mathrm{°aC}\)
\(V_{\mathrm{2}}\) = ?
Because we are looking for the volume change caused by a temperature change at constant pressure, this is a job for Charles’s law. Taking
\(\ce{V1}\) and
\(T_{\mathrm{1}}\) as the initial values,
\(T_{\mathrm{2}}\) as the temperature at which the volume is unknown and
\(\ce{V2}\) as the unknown volume we have:
\(\dfrac{V_{\mathrm{1}}}{T_{\mathrm{1}}} = \dfrac{V_{\mathrm{2}}}{T_{\mathrm{2}}}\)
Rearranging and solving gives:
\(V_{\mathrm{2}}\) \(= V_{\mathrm{1}} \cdot \dfrac{T_{\mathrm{2}}}{T_{\mathrm{1}}}\)
\(\ \ \ =0.300\ \mathrm{L} \cdot \dfrac{303.\ \mathrm{K}}{283.\ \mathrm{K}}\)
\(\ \ \ =0.300\ \mathrm{L} \cdot 1.0706\)
\(\ \ \ =0.321\ \mathrm{L}\)
This answer supports our expectation from Charles’s law, namely, that raising the gas temperature (from
283 K to
303 K) at a constant pressure will yield an increase in its volume (from
0.300 L to
0.321 L).