What happens to the pressure inside a rigid container if the amount of nitrogen gas increases from 0.500 mole to 0.750 mol? The original pressure is 98.0 kP

Question

What happens to the pressure inside a rigid container if the amount of nitrogen gas increases from 0.500 mole to 0.750 mol? The original pressure is 98.0 kPa, and the temperature remains constant.

Answer 1

Hi,

 

Anytime you see a problem that involves gases, your first instinct should be to use some form of the ideal gas law (PV = nRT), where P = pressure, V = volume of the container, n = moles, R = gas constant, and T = temperature.

 

First, realize that the number of moles of gas is increasing; therefore, we expect the pressure to increase (more gas molecules can collide with the container than before, resulting in a pressure increase). This can best be seen by solving the ideal gas law for pressure:

 

P= nRT/V

 

Let's write this out two times. One for the case of 0.500 moles (n₁) at 98.0 kPa (P₁), and the other for the case of 0.750 moles (n₂) at a new pressure (P₂). RT/V does not need subscripts since it is constant in this problem.

 

P₁ = n₁RT/V                        and    P₂ = n₂RT/V                      

 

Now, divide the first equation by the second equation to get:

 

P₁/P₂ = n₁/n₂     

 

Solving for P₂ gives: 

 

P₂ = P₁n₂/n₁ = (98.0 kPa)(0.750 mol)/(0.500 mol) = 147 kPa

 

Note that the pressure did increase (as we predicted) from 98.0 kPa to 147 kPa.