Question

What is the pressure exerted by 0.122 mol oxygen gas in a 1.50-L in a container at room temperature

Answer 1

Answer:

The answer to your question is P = 1.95 atm

Explanation:

Data

Pressure = P = ?

number of moles = n = 0.122

Volume = V = 1.5 l

Temperature = T = 20°C

Constant of ideal gases = 0.082 atm l/mol°K

Process

1.- Convert temperature to °K

T = 20 + 273

T = 293°K

2.- Write the formula of ideal gases

       PV = nRT

-Solve for P

      P = nRT/V

3.- Substitution

     P = (0.122 x 0.082 x 293) / 1.5

4.- Simplification

     P = 2.93 / 1.5

5.- Result

     P = 1.95 atm