Question

The gas in a closed container has a pressure of `3.00 * 10^2` `kPa` at `30°C` `(303 K)`. What will the pressure be if the temperature is lowered to `-172°C` `(146 K)`?

Answer 1

The final pressure will be 145 kPa.

Explanation:

This is an example of Gay-Lussac's law, which states that the pressure of a given gas is directly proportional to its Kelvin temperature, when held at a constant volume. The equation that represents this law is `P_1/T_1=P_2/T_2`.

Given/Known
`P_1=3.00xx10^2"kPa"`
`T_1="303 K"`
`T_2=-127^@"C+273.15=146 K"` **

Unknown
`P_2`

Solution
Rearrange the equation to isolate `P_2`. Substitute the given/known values into the equation and solve.

`P_1/T_1=P_2/T_2`

`P_2=(P_1T_2)/T_1`

`P_2=(3.00xx10^2"kPa" xx 146cancel"K")/(303cancel"K")="145 kPa"`

**Kelvin temperature changed to 146 K, which is `(-127^@ "C"+273.15)`

Excellent Resource for the gas laws: http://chemistry.bd.psu.edu/jircitano/gases.html