A gas is held at 3.8 atm and 500 K. If the pressure is then decreased to 1.2 atm, what will the new temperature be?

1 Answer
Mar 21, 2017

Answer:

#T_2 = 157.89 K#

#T_2 = -115.10 ^0C#

Explanation:

By changing the physical conditions like pressure, temperature and volume of the gas, the moles of the gas remain constant.
# n_1=n_2#

According to the ideal gas equation,

#(P_1V_1)/(RT_1) = (P_2V_2)/(RT_2) #

The volume of the process is also constant . As, according to the Avogadro's law,
#V alpha n#

So, #V_1=V_2#

#(P_1)/(T_1) = (P_2)/(T_2) #

#3.8/500 = 1.2/ T_2#

#T_2 =1.2/3.8 xx500#

#T_2 = 157.89 K#

#T_2 = -115.10 ^0C#

So, the new temperature will be #2280 K#.