Question

A sample of a gas in a rigid container has an initial pressure of 1.049 kPa and an initial temperature of 7.39 K. The temperature is increased to 30.70 K. What is the new pressure? Show your work in the work space below.

Answer 1

Answer: 4.358 kPa

Explanation:

The gas is contained within a rigid container, so the volume of the gas is constant. Therefore, we can use Gay-Lussac's law, which states that:

"for a gas kept at constant volume, the pressure and the absolute temperature are directly proportional"

In formulas:

$\frac{P_1}{T_1}=\frac{P_2}{T_2}$

where:

$P_1 = 1.049 kPa$ is the initial pressure

$T_1 = 7.39 K$ is the initial temperature

$P_2$ is the final pressure

$T_2 = 30.70 K$ is the final temperature

Substituting the numbers into the equation, we find

$P_2 = P_1 \frac{T_2}{T_1}=(1.049 kPa)\frac{30.70 K}{7.39 K}=4.358 kPa$