The gas inside of a container exerts #48 Pa# of pressure and is at a temperature of #270 ^o K#. If the pressure in the container changes to #60 Pa# with no change in the container's volume, what is the new temperature of the gas?

1 Answer
Mar 2, 2016

Answer:

#337.5K#

Explanation:

Recall that Gay-Lussac's Law states that pressure is directly proportional to temperature. The formula that represents this is:

#color(brown)(P_1)/color(blue)(T_1)=color(purple)(P_2)/color(orange)(T_2)#

where:
#color(brown)(P_1)=#initial pressure
#color(blue)(T_1)=#inital temperature (Kelvin)
#color(purple)(P_2)=#final pressure
#color(orange)(T_2)=#final temperature (Kelvin)

Finding the Final Temperature
#1#. Since all other values have been given, the only variable you need to solve for is #T_2#, the final temperature. Thus, start by rearranging the formula to solve for #T_2#.

#P_1/T_1=P_2/T_2#

#T_2=(P_2T_1)/P_1#

#2#. Substitute your known values into the rearranged formula.

#color(purple)(P_2=60Pa)color(white)(XXX)color(blue)(T_1=270K)color(white)(XXX)color(brown)(P_1=48Pa)#

#T_2=((60Pa)(270K))/((48Pa))#

#3#. Solve for #T_2#.

#color(green)(T_2=337.5K)#

#:.#, the new temperature of the gas is #337.5K#.