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

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 = 2.79 atm

`P_2` = final pressure of gas = ?

`T_1` = initial temperature of gas = 273K

`T_2` = final temperature of gas = 315 K

`(2.79)/(273)=(P_2)/(315)`

`P_2=3.22atm`

Thus the resultant pressure is 3.22 atm