Question

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

Answer 1

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 = "0 Celsius"`
`P_1="1 atm"`

A change in temperature causes a change in the pressure:
`T_2 = "-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:

`(P_1)/(T_1)=(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 = "0 Celsius" + 273.15 = 273.15 K`

`T_2 = "-25 Celsius" + 273.15 = 248.15 K`

Plugging in the values to find `P_2`:

`("1 atm")/("273.15 K")=(P_2)/("248.15 K")`

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