Determine the pressure exerted by 0.352 mol of oxygen gas in a 1.75 L container at room temperature (25 degrees Celsius). Show all your work!

P = 4.92 atm

Explanation:

Given data:

Number of mole of oxygen = 0.352 mol

Volume of gas = 1.75 L

Temperature = 25°C

Pressure exerted by gas = ?

Solution:

The given problem will be solve by using general gas equation,

PV = nRT

P= Pressure

V = volume

n = number of moles

R = general gas constant = 0.0821 atm.L/ mol.K  

T = temperature in kelvin

Now we will convert the temperature.

25+273 = 298 K

P × 1.75 L = 0.352 mol  × 0.0821 atm.L/ mol.K   ×298 K

P = 8.61 atm.L /  1.75 L

P = 4.92 atm

points

explanation: