Question

A gas confined in a rigid container exerts a pressure of 9.9 atm at a temperature of `17^@"C"`. What will the pressure be in atm in the container if the gas is cooled to `-23^@"C"`?

Answer 1

The pressure at `"250 K"` `(-23^@"C")` will be `"8.5 atm".`

Explanation:

The variables are pressure and temperature, so this is an example of Gay-Lussac's law, which states that the pressure of a given amount of gas held at constant volume is directly proportional to the Kelvin temperature . This means that if the pressure increases, the temperature also increases, and vice versa. The equation for this law is:

`P_1/T_1=P_2/T_2`,

where:

`P_1` and `P_2` are the first and second pressures, respectively, and `T_1` and `T_2` are the first and second temperatures, respectively.

Organize the data:

Known

`P_1="9.9 atm"`

`T_1="17"^@"C"+273.15="290 K"`

`T_2="-23"^@"C"+273.15="250 K"`

Unknown

`P_2`

Solution

Rearrange the equation to isolate `P_2`. Plug in the known values and solve.

`P_2=(P_1T_2)/T_1`

`P_2=("9.9 atm"xx250color(red)cancel(color(black)("K")))/(290color(red)cancel(color(black)("K")))="8.5 atm"`