Question

The gas inside of a container exerts `48 Pa` of pressure and is at a temperature of `320 ^o K`. If the pressure in the container changes to `64 Pa` with no change in the container's volume, what is the new temperature of the gas?

Answer 1

The new temperature is `=426.7K`

Explanation:

We apply Gay Lussac's Law

`P_1/T_1=P_2/T_2`

The initial pressure is `P_1=48Pa`

The initial temperature is `T_1=320K`

The final pressure is `P_2=64Pa`

The final temperature is

`T_2=P_2/P_1*T_1`

`=64/48*320`

`=426.7K`

Answer 2

use gay-lussac's law .

Explanation:

gay-lusaac's law gives you the relation between pressure and temperature of a gas inside a container with a fixed volume :
`P_1/T_1 = P_2/T_2`
`P_1` is the pressure exerted by a gas at temperature `T_1` and `P_2` is the pressure exerted by the gas at temperature `T_2` .
now, substituting the values :
`48/320 =64/T`
`T= 426.67 k`