A container with a volume of #7 L# contains a gas with a temperature of #360^o C#. If the temperature of the gas changes to #540 ^o K# without any change in pressure, what must the container's new volume be?

Question

A container with a volume of #7  L# contains a gas with a temperature of #360^o   C#.  If the temperature of the gas changes to #540     ^o    K# without any change in pressure, what must the container's new volume be?

Nathan Ashcroft · Accepted Answer

#V_f=5.97L# #"initial volume of gas :"v_i=7L# #"initial temperature of gas :"T_i=360 ^o=360+273=633 ^o K# #"final temperature of gas :"540 ^o K# #V_i/T_i=V_f/T_f# #7/633=V_f/540# #V_f=(540*7)/633# #V_f=5.97L#

Axel Acevedo · Answer

undefined To find the new volume of the gas, we can use the combined gas law formula:

$ \frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2} $

Given:
Initial volume ($ V_1 $) = 7 L
Initial temperature ($ T_1 $) = 360°C = 633 K
Final temperature ($ T_2 $) = 540 K

Since the pressure remains constant, we can rearrange the equation to solve for the new volume ($ V_2 $):

$ V_2 = \frac{P_1V_1T_2}{P_2T_1} $

Since the pressure ($ P_1 $ and $ P_2 $) is not given, we can assume it cancels out. Therefore, we can simplify the equation to:

$ V_2 = V_1 \times \frac{T_2}{T_1} $

Now, plug in the given values:

$ V_2 = 7 \, \text{L} \times \frac{540 \, \text{K}}{633 \, \text{K}} $

$ V_2 \approx 5.98 \, \text{L} $

So, the new volume of the gas is approximately 5.98 L.