# 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 thepressure. We can use#"atm"# .#V# is thevolumein#"L"# .#n# is the#bb("mol")# sof gas.#R = "0.082057 L"cdot"atm/mol"cdot"K"# is theuniversal gas constantfor when you use pressure units of#"atm"# (you do use a different one depending on your units of pressure and volume).#T# is thetemperaturein#"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,