Question

You fill a rigid steel cylinder that has a volume of 20. 0 L with 10. Moles of nitrogen gas at 313 ºK. What is the final pressure in kPa in the cylinder?

Answer 1

Assuming ideal gas behavior, we can use the ideal gas law:

PV = nRT

where P is the pressure, V is the volume, n is the number of moles, R is the gas constant, and T is the temperature in Kelvin.

First, we need to convert the temperature from Celsius to Kelvin:

T = 313 ºK

The gas constant R is 8.31 J/(mol·K), but we need to use a value in units of kPa·L/(mol·K) for the pressure and volume units given:

R = 8.31 kPa·L/(mol·K)

Plugging in the given values, we have:

P × 20.0 L = 10.0 mol × 8.31 kPa·L/(mol·K) × 313 K

Solving for P, we get:

P = (10.0 mol × 8.31 kPa·L/(mol·K) × 313 K) / 20.0 L

P = 1293 kPa

Therefore, the final pressure in the cylinder is 1293 kPa.

Answer 2

The quantity of nitrogen the cylinder contains is 4477.8 g

Step-by-step explanation:

hope this helps u out:))