The gas inside of a container exerts #48 Pa# of pressure and is at a temperature of #320 ^o K#. If the pressure in the container changes to #64 Pa# with no change in the container's volume, what is the new temperature of the gas?

2 Answers
Apr 15, 2017

Answer:

The new temperature is #=426.7K#

Explanation:

We apply Gay Lussac's Law

#P_1/T_1=P_2/T_2#

The initial pressure is #P_1=48Pa#

The initial temperature is #T_1=320K#

The final pressure is #P_2=64Pa#

The final temperature is

#T_2=P_2/P_1*T_1#

#=64/48*320#

#=426.7K#

Apr 15, 2017

Answer:

use gay-lussac's law .

Explanation:

gay-lusaac's law gives you the relation between pressure and temperature of a gas inside a container with a fixed volume :
# P_1/T_1 = P_2/T_2 #
#P_1 # is the pressure exerted by a gas at temperature #T_1# and #P_2# is the pressure exerted by the gas at temperature # T_2 # .
now, substituting the values :
#48/320 =64/T #
# T= 426.67 k #