Answer for a question below

Question

A sample of 3.50 mol of gas in a 10.00 L container is at 45.0 °C. What is the pressure of the gas?

Answer 1

PV=nRT

Where:

P is the pressure (in atmospheres, atm)

V is the volume (in liters, L)

n is the number of moles of the gas

R is the ideal gas constant (0.0821 L⋅atm/K⋅mol0.0821L⋅atm/K⋅mol)

T is the temperature (in Kelvin, K)

You have the following information:

Moles (n) = 3.50 mol

Volume (V) = 10.00 L

Temperature (T) = 45.0 °C = 318.15 K (Note: Temperature must be in Kelvin)

Now, plug these values into the ideal gas law and solve for P.

⋅10.00 L=(3.50 mol)⋅(0.0821 L⋅atm/K⋅mol)⋅318.15 KP⋅10.00L=(3.50mol)⋅(0.0821L⋅atm/K⋅mol)⋅318.15K

Now, calculate the pressure (P):

=(3.50 mol)⋅(0.0821 L⋅atm/K⋅mol)⋅318.15 K10.00 LP=10.00L(3.50mol)⋅(0.0821L⋅atm/K⋅mol)⋅318.15K​

P ≈ 8.95 atm

So, the pressure of the gas is approximately 8.95 atmospheres.