Question

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.

Answer 1

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 `V`, `n`, and `R` are the same across both equations because the container is closed (constant `n`) and rigid (constant `V`).

(Obviously, the universal gas constant is constant.)

`P_1V = nRT_1`
`P_2V = nRT_2`

Therefore, we can divide these equations to solve for `T_2`.

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

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

`=> color(blue)(T_2) = T_1xx(P_2)/(P_1)`

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

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

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