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

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

Answer 2

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)

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

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.

$\frac{7.58 kPa}{639K}=\frac{P_2}{311K}$

$P_2=3.69kPa$

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