Question

The gas inside of a container exerts `6 Pa` of pressure and is at a temperature of `120 ^o C`. If the temperature of the gas changes to `180 ^oK` with no change in the container's volume, what is the new pressure of the gas?

Answer 1

`"New pressure of the gas is 2.75Pa"`

Explanation:

`"We should use the Gay-Lussac's law to solve this problem."`

`P_i/T_i=P_f/T_f`

`P_i:"The initial pressure of gas "P_i=6Pa`
`T_i:"The initial temperature of gas "T_i=120^o C`

`"We should change "^o C " to " ^o K "(Kelvin)"`

`T_i=120+273=393^o K`

`T_f:"Final temperature of gas "T_f=180^o K`

`P_f:"Final pressure of gas "P_f=?`

`6/393=P_f/180`

`6*180=393*P_f`

`P_f=(6*180)/393`

`P_f=2.75Pa`