Question

The gas inside of a container exerts `42 Pa` of pressure and is at a temperature of `420 ^o K`. If the pressure in the container changes to `72 Pa` with no change in the container's volume, what is the new temperature of the gas?

Answer 1

`"final temperature is "720 ^o K`

Explanation:

`"As the volume of the container does not change, the increase of"`
`"the gas pressure causes the temperature to increase."`

`"We need to use the Gay-Lussac law to find the temperature."`

`P_i=42 Pa("initial pressure of the gas")`
`P_f=72 Pa("final pressure of the gas")`
`T_1=420 ^o K("initial temperature of the gas")`
`T_f=?("final temperature of the gas")`

`P_i/T_i=P_f/T_f`

`cancel(42)/cancel(420)=72/T_f`

`"let's divide both numerator and denominator by 42 "`

`1/10=72/T_f`

`1*T_f=72*10`

`T_f=720 ^oK`