# How do you calculate the temperature of the gas at this new pressure?

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A sample of an ideal gas is sealed in a container whose volume cannot change. Initially it has a pressure of 2.5 x 10^5 Pa and its temperature -20 oC.The gas is cooled until the pressure is reduced to 1.0 x 10^5 Pa.

A sample of an ideal gas is sealed in a container whose volume cannot change. Initially it has a pressure of 2.5 x 10^5 Pa and its temperature -20 oC.The gas is cooled until the pressure is reduced to 1.0 x 10^5 Pa.

##### 1 Answer

Since you are talking about volume, pressure, and temperature, we can naturally use the **ideal gas law**.

#\mathbf(PV = nRT)# where:

#P# is thepressurein#"Pa"# .#V# is thevolumein#"L"# .#n# is#"mol"# s of gas, which hasnotchanged throughout the process.#R# is theuniversal gas constant. We won't need to use this because it is a constant throughout the process.#T# is thetemperaturein #"K".

You are told that the volume cannot change, so the volume stays *constant*. Note your variables:

#P_1 = 2.5xx10^5 "Pa"#

#P_2 = 1.0xx10^5 "Pa"#

#T_1 = -20^@ "C" = "253.15 K"#

#T_2 = ?#

So, set up your equations using these variables. We know that **closed** (constant **rigid** (constant

(Obviously, the universal gas *constant* is constant.)

#P_1V = nRT_1#

#P_2V = nRT_2#

Therefore, we can divide these equations to solve for

#(P_1cancel(V))/(P_2cancel(V)) = (cancel(nR)T_1)/(cancel(nR)T_2)#

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

#= ("253.15 K")xx(1.0xx10^5 cancel"Pa")/(2.5xx10^5 cancel"Pa")#

#=# #color(blue)("101.26 K")#

#=# #color(blue)(-171.89^@ "C")#