A container with a volume of #12 L# contains a gas with a temperature of #240^o C#. If the temperature of the gas changes to #450 ^o K# without any change in pressure, what must the container's new volume be?

Question

A container with a volume of #12  L# contains a gas with a temperature of #240^o   C#.  If the temperature of the gas changes to #450     ^o    K# without any change in pressure, what must the container's new volume be?

Emily Abbate · Accepted Answer

#"The final volume is "10.53L# #"The given data:"# #V_i:"Initial volume of gas ;"V_i=12L# #V_f:"Final volume of gas ;"V_f=?# #T_i:"Initial temperature of gas ;"T_i=240^o C# #T_i=240+273=513^o K# #T_f:"Final temperature of gas ;"T_f=450^o K# #V_i/T_i=V_f/T_f# #12/513=V_f/450# #V_f=(12*450)/513# #V_f=5400/450# #V_f=10.53L" rounded to two decimal place"#

Elena Albarran · Answer

undefined To find the new volume, we can use the combined gas law, which states that $ \frac{{P_1V_1}}{{T_1}} = \frac{{P_2V_2}}{{T_2}} $, where $ P $ is the pressure, $ V $ is the volume, and $ T $ is the temperature in Kelvin. Since the pressure is constant, we can simplify the formula to $ \frac{{V_1}}{{T_1}} = \frac{{V_2}}{{T_2}} $. Plugging in the values, we get $ \frac{{12}}{{240 + 273}} = \frac{{V_2}}{{450}} $. Solving for $ V_2 $, we get $ V_2 = \frac{{12 \times 450}}{{240 + 273}} $. Calculating this gives $ V_2 \approx 16.67 \, \text{L} $.