How do you calculate the temperature of the gas at this new pressure?
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 the pressure in#"Pa"# .#V# is the volume in#"L"# .#n# is#"mol"# s of gas, which has not changed throughout the process.#R# is the universal gas constant. We won't need to use this because it is a constant throughout the process.#T# is the temperature in #"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
(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")#