Question

A sample of a gas has a pressure of `"0.370 atm"` at a temperature of `"5.00"^@"C"`. What will the pressure be if the temperature is lowered to `"0"^@"C"`?

Answer 1

The pressure at `"273.15 K"` `(0^@"C")` will be `"0.363 atm"`.

Explanation:

Gay-Lussac's gas law will be used to solve this problem. It states that the pressure of a gas is directly proportional to its Kelvin temperature, as long as mass and volume are held constant. This means that if the volume increases, so does the temperature and vice-versa. The equation for this law is:

`P_1/T_1=P_2/T_2`

The given data has temperature in degrees Celsius, so it needs to be converted to Kelvins. This is done by adding `"273.15"` to the Celsius temperature.

Organize your data:

Given

`P_1="0.370 atm"`

`T_1="5.00"^@"C" + 273.15="278.15 K"`

`T_2="0.00"^@"C + 273.15 ="273.15 K"`

Unknown:

`P_2`

Solution
Rearrange the equation to isolate `P_2`. Insert the given data and solve.

`P_2=(P_1T_2)/(T_1)`

`P_2=(0.370color(white)(.)"atm"xx273.15color(red)cancel(color(black)("K")))/(278.15color(red)cancel(color(black)("K")))="0.363 atm"` (rounded to three significant figures)