if 3.0 moles of nitrogen gas are collected in a 35.0 liter container at 20 C what would be the the pressure exerted on the container in atmospheres

Answer 1

PV = nRT

where P = pressure, V = volume, R is a constant = 8.314 J/mol*K, and T is temperature in degrees Kelvin, n is number of moles

degrees K = 273 + degrees C

1 liter = 1000 cm^3 = 1000 cm^3* (1 m / 100 cm)^3 = 1000 * (1/1000000) = 1/1000
= 0.001 m^3

35 L = 0.0035 m^3

P*0.0035 = 3*8.314*293

Solve for P

Answer 2

I believe 35.0L is 0.035 m^3; i.e.,
0.001 x 35.0L = 0.0350 m^3

Answer 3

.035