Question

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

Answer 1

The temperature will be `"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 `"100 kPa"`. `larr` 100 kiloPascals

We need to convert `"mmHg"` to `"kPa"`.

`1"kPa"="7.5006 mmHg"`

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

Organize your data:

Known

`P_1="93.2 kPa"`

`T_1="40"^@"C" + 273.15="313 K"`

`P_2="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=(P_2T_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 `"336 K"`.