# The temperature of 200.0 mL of a gas originally at STP is changed to -25C at constant volume. Calculate the pressure of the gas in atm? (0.91 atm)

## gay-lussac's law

Aug 1, 2018

0.91 atm

#### Explanation:

At standard temperature and pressure ( STP ), the system has a temperature of 0 Celsius and pressure which is equal to the atmosphere, 1 atm.

${T}_{1} = \text{0 Celsius}$
${P}_{1} = \text{1 atm}$

A change in temperature causes a change in the pressure:
${T}_{2} = \text{-25 Celsius}$
P_2 = ?

The volume of the gas remains constant at 200.0 mL, while its pressure and temperature changed. We will be able to calculate the pressure, ${P}_{2}$, after the temperature change using the Gay-Lusaac's law:

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

Before plugging in the values to the formula, we need to first convert the temperatures from Celsius to Kelvin:

${T}_{1} = \text{0 Celsius} + 273.15 = 273.15 K$

${T}_{2} = \text{-25 Celsius} + 273.15 = 248.15 K$

Plugging in the values to find ${P}_{2}$:

$\left(\text{1 atm")/("273.15 K")=(P_2)/("248.15 K}\right)$

P_2 = ("1 atm")/("273.15 K") xx "248.15 K" = "0.91 atm"