The pressure of a sample of argon at 20°C is decreased from 720 mmHg to 360 mmHg. What was the final temperature of the argon gas?

1 Answer
Jun 25, 2016

Answer:

#T_2 = 146.5 K#

or

#T_2 = -126.5^oC#

Explanation:

For this solution we will use the Gay-Lussac Law relating pressure and temperature in gases. #P_1/T_1 = P_2/T_2#

For this case the values are

#P_1 = 720 mmHg#
#T_1 = 20^oC or 293K#
#P_2 = 360 mmHg#
#T_1 = ?K#

#P_1/T_1 = P_2/T_2#

#(720mmHg)/(293K) = (360mmHg)/T_2#

We can invert both sides to put the missing variable in the numerator.

#(293K)/(720mmHg) = T_2/(360mmHg)#

#(293K)/(720cancel(mmHg))(360cancel(mmHg)) = T_2/(cancel(360mmHg))(cancel(360mmHg))#

#T_2 = 146.5 K#

or

#T_2 = -126.5^oC#