Question

0.650 moles of O₂(g) are added to 1.00 L flask and the internal pressure ismeasured at 20.00 atm. What is the temperature of the gas under theseconditions (in *C)? ________ °C?

Answer 1

1) List the known and unknown quantities.

Sample: O2

Amount of substance: 0.650 mol O2.

Volume: 1.00 L.

Pressure: 20.00 atm.

Ideal gas constant: 0.082057 L * atm * K^(-1) * mol^(-1)

Temperature: unknown.

2) Set the equation.

Ideal gas law.

`PV=nRT`

3) Plug in the known quantities and solve for T.

`(20.00\text{ }atm)(1.00\text{ }L)=(0.650\text{ }mol\text{ }O_2)(0.082057\text{ }L*atm*K^(-1)*mol^(-1))(T)`
`T=\frac{(20.00atm)(1.00L)}{(0.650\text{ }mol\text{ }O_2)(0.082057\text{ }L*atm*K^(-1)*mol^(-1))}`
`T=374.97\text{ }K`

4) Convert K to ºC.

`ºC=K-273.15`
`ºC=374.94-273.15`
`ºC=101.81\text{ }ºC`

The temperature of the gas under these conditions is 101.81ºC.

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