When the volume #V_1# of a gas is halved at constant pressure, what is its new temperature if it began at #0^@ "C"#?
1 Answer
I get
Any gas problem where you are not told what the gas is, or you are not told to use a real gas, can be assumed to be an ideal gas problem.
Therefore, you can use the ideal gas law:
#PV = nRT# where:
#P# is the pressure . We can use#"atm"# .#V# is the volume in#"L"# .#n# is the#bb("mol")# s of gas.#R = "0.082057 L"cdot"atm/mol"cdot"K"# is the universal gas constant for when you use pressure units of#"atm"# (you do use a different one depending on your units of pressure and volume).#T# is the temperature in#"K"# .
You are told that you are at constant pressure, so
Notice how if pressure is constant, you only have two variables changing: volume and temperature.
- Pressure is constant.
- The mols of gas are constant.
- The universal gas constant is... constant, no surprise.
You are then left with the following two equations:
#PV_1 = nRT_1#
#PV_2 = nRT_2#
When you divide these, you get:
#V_2/V_1 = T_2/T_1#
So, to solve for the new temperature, you were told that the volume was halved. Therefore,
#(1/2cancel(V_1))/(cancel(V_1)) = T_2/T_1#
#-> T_2 = 1/2T_1#
Note that
#-> color(blue)(T_2) = 1/2("600.15 K")#
#=# #color(blue)("300.08 K")#
Or,