Question

5.00 L of a gas is collected at 22.0 'C and 745.0 mmHg . When the temperature increases to 35'C what is the new pressure

Answer 1

The new pressure is 777.926 mmHg.

Step-by-step explanation:

We use the ideal gas equation

`PV = nrT`.

where
`P, V , n,` and
`T` are the pressure, volumes, moles, and temperature of the gas respectively.
`r` is the gas constant
`r = 8.3145\: j*mol^(-1)*K^(-1)`.

Initially, for the gas

`V = 5L = 0.005m^3`

`P = 745.0 \:mmHg = 99325.2\: Pa`

`T = 22^oC = 295\:K.`

therefore, the number of moles
`n` is

`(99325.2\:Pa)(0.005m^3) = n( 8.3145\: j*mol^(-1)*K^(-1))(295K)`

`n = ((99325.2\:Pa)(0.005m^3))/(( 8.3145\: j*mol^(-1)*K^(-1))(295K))`

`n= 0.2025`

Now, when the temperature rises to 35°C, (assuming constant volume) we have

`P = (0.2025mol*8.3145\: j*mol^(-1)*K^(-1)308K )/(0.005m^3)`

`P = 103715.1 \: Pa`

or in mmHg this is

`\boxed{P = 777.926\:mmHg}`