# A sample of gas at "31"^@"C" has a pressure of "745 Torr". What is the pressure if the temperature is increased to "78"^@"C"?

Apr 9, 2017

The final pressure will be $\text{860. torr}$.

#### Explanation:

This is an example of Gay-Lussac's law , which states that the pressure of a gas with a constant volume and amount, is directly proportional to the temperature in Kelvins. This means that if the pressure is increased, the temperature will increase, and vice-versa.

The equation that is used for this law is:

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

where ${P}_{1}$ is the initial pressure, ${P}_{2}$ is the final pressure, ${T}_{1}$ is the initial temperature, ${T}_{2}$ is the final temperature.

Write what is known.
${P}_{1} = \text{745 torr}$
${T}_{1} = \text{31"^@"C" + 273.15=304 "K}$
${T}_{2} = \text{78"^@"C" + 273.15=351 "K}$

Write what is unknown: ${P}_{2}$

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

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

P_2=(745"torr"xx351color(red)cancel(color(black)("K")))/(304color(red)cancel(color(black)("K")))="860. torr" (rounded to three significant figures)

Apr 9, 2017

${P}_{2} = 860$ torr

#### Explanation:

With a constant container volume, Charles' Law becomes simply:
${P}_{1} / {T}_{1} = {P}_{2} / {T}_{2}$
${P}_{2} = {P}_{1} \cdot {T}_{2} / {T}_{1}$ ; ${P}_{2} = 745 \cdot \frac{351}{304}$ ; ${P}_{2} = 860$ torr
$\frac{{P}_{1} \cdot {V}_{1}}{T} _ 1 = \frac{{P}_{2} \cdot {V}_{2}}{T} _ 2$
Where ${P}_{1} {,}_{2}$ are pressures – units don't matter in this case as long as they are consistent, because this is a ratio.
${V}_{1} {,}_{2}$ are the corresponding volumes in Liters
${T}_{1} {,}_{2}$ are the temperatures in degrees Kelvin