A 6.69 L cylinder contains 1.62 mol of gas A and 4.00 mol of gas B, at a temperature of 27.3 °C. Calculate the partial pressure of each gas in the cylinder. Assume ideal gas behavior.

The partial pressure of gas A is 5.97 atm

The partial pressure of gas B is 14.73 atm

Explanation:

From the ideal gas equation

PV = nRT

P = nRT/V

n is total moles of gas mixture = 1.62 + 4 = 5.62 moles

R = 82.057×10^-3 L.atm/gmol.K

T = 27.3°C = 27.3+273 = 300.3 K

V = 6.69 L

P (total pressure in the cylinder) = 5.62×82.057×10^-3×300.3/6.69 = 20.7 atm

Partial pressure of gas A = 1.62/5.62 × 20.7 atm = 5.97 atm

Partial pressure of gas B = 4/5.62 × 20.7 atm = 14.73 atm

not too sure thinking all accelerations are pointed down