# A gas has a pressure of 699.0 mm Hg at 40.0°C. What is the temperature at standard pressure?

Jul 6, 2017

The temperature will be $\text{336 K}$.

#### Explanation:

This is a pressure-temperature gas problem. This means that it involves Gay-Lussac's law, which states that the pressure of a given amount of a gas, held at constant volume, varies directly with its temperature. This means that if the pressure increases, so does the temperature, and vice-versa. The equation used to solve this problem is:

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

Before we go further, we need to determine what standard pressure is, and we need to convert the Celsius temperature to Kelvin temperature by adding $273.15$ to the Celsius temperature.

Standard pressure is $\text{100 kPa}$. $\leftarrow$ 100 kiloPascals

We need to convert $\text{mmHg}$ to $\text{kPa}$.

$1 \text{kPa"="7.5006 mmHg}$

699.0color(red)cancel(color(black)("mmHg"))xx(1"kPa")/(7.5006color(red)cancel(color(black)("mmHg")))="93.2 kPa"

Known

${P}_{1} = \text{93.2 kPa}$

${T}_{1} = \text{40"^@"C" + 273.15="313 K}$

${P}_{2} = \text{100 kPa}$

Unknown

T_2=?

Solution

Rearrange the equation above to isolate ${T}_{2}$. Insert the given data into the new equation and solve.

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

T_2=(100color(red)cancel(color(black)("kPa"))xx313"K")/(93.2color(red)cancel(color(black)("kPa")))="336 K"

If you need to convert the temperature in Kelvins back to the Celsius temperature, subtract $273.15$ from $\text{336 K}$.