Question

A 30.0 L sample of nitrogen inside a rigid, metal container at 

20.0 °C is placed inside an oven whose temperature is 50.0 °C. 

The pressure inside the container at 20.0 °C was at 3.00 atm. 

What is the pressure of the nitrogen after its temperature is 

increased to 50.0 °C? 

Answer 1

Since the volume of the container remains constant, it may not be involved in the calculation, and Gay-Lussac's law is applicable:

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

Temperatures should be converted to Kelvins for calculation:

T₁ = 20.0^oC = 273 + 20 = 293 K

T₂ = 50.0^oC = 273 + 50 = 323 K

Therefore,

$P_2=\frac{P_1T_2}{T_1}=\frac{3.00\ atm\times323\ K}{293\ K}=3.31\ atm$

Answer: 3.31 atm