Question

The total pressure in an 12.7L automobile tire is 42.0psi at `16.0^@"C"`. How much does the pressure in the tire rise if it warms to a temperature of `39.0^@"C"` and the volume remains at 12.7L ?

Answer 1

The pressure will increase `3.30"psi"` to `45.3"psi"`.

Since the volume remains constant, you can use Gay-Lussac's law, which states that the pressure of a given amount of gas held at constant volume is directly proportional to the Kelvin temperature. http://chemistry.bd.psu.edu/jircitano/gases.html

The equation for Gay-Lussac's law is `P_1/T_1=P_2/T_2` .

Given:
`P_1=42.0"psi"`
`T_1=16.0^"o""C"+273.15=289"K"`
`T_2=39.0^"o""C"+273.15=312"K"`

Unknown:
`P_2`

Solution: Substitute the given values into the equation and solve the equation for `P_2`.

`P_2=(P_1T_2)/T_1=(42.0"psi"*312cancel"K")/(289cancel"K")=45.3"psi"`

Increase in pressure = `45.3"psi"-42.0"psi"=3.30"psi"`