# The gas inside of a container exerts 16 Pa of pressure and is at a temperature of 430 ^o K. If the temperature of the gas changes to 120 ^oC with no change in the container's volume, what is the new pressure of the gas?

Mar 1, 2018

$14.63 \text{Pa}$

#### Explanation:

According to Gay-Lussac's Law, when volume is constant,

$P \propto T$, and so:

$\frac{P}{T} = k$, and it so follows that:

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

Here, we have:

${P}_{1} = 16 \text{Pa}$

${T}_{1} = 430 \text{K}$

${T}_{2} = 120 + 273.15 = 393.15 \text{K}$

We rearrange to solve for ${P}_{2}$:

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

Inputting:

${P}_{2} = \frac{16 \cdot 393.15}{430}$

${P}_{2} = 14.63 \text{Pa}$

Mar 1, 2018

The new pressure is $= 14.62 P a$

#### Explanation:

Apply Gay Lussac's Law

${P}_{1} / {T}_{1} = {P}_{2} / {T}_{2}$ at $\text{constant volume}$

The initial pressure is ${P}_{1} = 16 P a$

The initial temperature is ${T}_{1} = 430 K$

The final temperature is ${T}_{2} = 120 + 273 = 393 K$

The final pressure is

${P}_{2} = {T}_{2} / {T}_{1} \cdot {P}_{1} = \frac{393}{430} \cdot 16 = 14.62 P a$