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.

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)

${P_1\times T_1}={P_2\times T_2}$

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  = ?

${1.00\times 373}={1.13\times T_2}$

$T_2=330K$

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