Question

What is the resultant pressure if 1.7 mol of ideal gas at 273 K and 2.79 atm in a closed container of constant volume is heated to 315 K? Answer in units of atm.

Answer 1

Answer: The resultant pressure is 3.22 atm

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)

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

where,

$P_1$ = initial pressure of gas  = 2.79 atm

$P_2$ = final pressure of gas  = ?

$T_1$ = initial temperature of gas  = 273K

$T_2$ = final temperature of gas = 315 K

$\frac{2.79}{273}=\frac{P_2}{315}$

$P_2=3.22atm$

Thus the resultant pressure is 3.22 atm