Question

When the volume `V_1` of a gas is halved at constant pressure, what is its new temperature if it began at `0^@ "C"`?

Answer 1

I get `"300 K"`, or `26.9^@ "C"`.

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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 `DeltaP = P_2 - P_1 = 0`, meaning `P_2 = P_1`. You are also given the initial temperature and volume.

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, `V_2 = 1/2V_1`. You then have:

`(1/2cancel(V_1))/(cancel(V_1)) = T_2/T_1`

`-> T_2 = 1/2T_1`

Note that `0^@ "C" + 273.15 = "273.15 K"`, so we'll use `"600.15 K"`.

`-> color(blue)(T_2) = 1/2("600.15 K")`

`=` `color(blue)("300.08 K")`

Or, `300.08 - 273.15 ~~ color(blue)(26.9^@ "C")`.