# If the initial temperature of an ideal gas at 2.250 atm is 62.0 degrees C, what final temperature would cause the pressure to be reduced to 1.750 atm?

Oct 22, 2017

The final temperature would be $\text{260.7 K}$ or $- {12.45}^{\circ} \text{C}$.

#### Explanation:

This question is an example of Gay-Lussac's law relating temperature and pressure. This law 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:

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

The temperature must be in Kelvins. To convert degrees Celsius to Kelvins, add $273.15$ to the Celsius temperature.

Organize the data:

Known

${P}_{1} = \text{2.250 atm}$

${T}_{1} = \text{62.0"^@"C" + 273.15="335.2 K}$

${P}_{2} = \text{1.750 atm}$

Unknown

${T}_{2}$

Solution

Rearrange the equation to isolate ${T}_{2}$. Insert the known data and solve.

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

T_2=(1.750"atm"xx335.2"K")/(2.250"atm")="260.7 K"

To convert Kelvins to degrees Celsius, subtract $273.15$ from the Kelvin temperature.

$260.7 - 273.15 = \text{-12.45"^@"C}$