Question

At 1.00 atmosphere pressure, a certain mass of a gas has a temperature of 100oC. What will be the temperature at 1.13 atmosphere pressure if the volume remains constant?

Answer 1

Answer: Final temperature of the gas will be 330 K.

Step-by-step explanation:

Gay-Lussac's Law: This law states that pressure is directly proportional to the temperature of the gas at constant volume and number of moles.

`P\propto T` (At constant volume and number of moles)

`{P_1* T_1}={P_2* T_2}`

where,

`P_1` = initial pressure of gas = 1.00 atm

`P_2` = final pressure of gas = 1.13 atm

`T_1` = initial temperature of gas =
`100^0C=(100+273)K=373K` K

`T_2` = final temperature of gas = ?

`{1.00* 373}={1.13* T_2}`

`T_2=330K`

Therefore, the final temperature of the gas will be 330 K.