# 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)?

Feb 22, 2016

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
${P}_{1} = 3.00 \times {10}^{2} \text{kPa}$
${T}_{1} = \text{303 K}$
${T}_{2} = - {127}^{\circ} \text{C+273.15=146 K}$ **

Unknown
${P}_{2}$

Solution
Rearrange the equation to isolate ${P}_{2}$. Substitute the given/known values into the equation and solve.

${P}_{1} / {T}_{1} = {P}_{2} / {T}_{2}$

${P}_{2} = \frac{{P}_{1} {T}_{2}}{T} _ 1$

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

**Kelvin temperature changed to 146 K, which is $\left(- {127}^{\circ} \text{C} + 273.15\right)$

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