Question

A sample of nitrogen gas has a pressure of 7.58 kPa at 639 K. What will be the pressure at 311 K if the volume does not change?

Answer 1

Answer: 3.69 kPa

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 = 7.58 kPa

`P_2` = final pressure of gas = ?

`T_1` = initial temperature of gas = 639 K

`T_2` = final temperature of gas = 311 K

Now put all the given values in the above equation, we get the final pressure of gas.

`(7.58 kPa)/(639K)=(P_2)/(311K)`

`P_2=3.69kPa`

Therefore, the final pressure of gas will be 3.69 kPa.

Answer 2

Over here you have to use the formula: P1/T1= P2/T2
P1= 7.58
T1= 639

P2= ?
T2= 311

7.58/639 = ?/311
= 3.689 kPa