At what pressure would 0.150 mole of nitrogen gas at -23.0oC occupy 8.90 L?
-23.o0c = 250 K
P=(0.150 mol)(8.31 KPA x L/ K x mol)(250 K) = P
is this right?
I know what to do in this question. My teacher showed me.
Your approach is correct, but there seems to be a small error in your calculation. Let me explain the correct way to solve this problem step by step.
To find the pressure, we can use the ideal gas law equation:
PV = nRT
Where:
P is the pressure (in units of kPa or atm)
V is the volume (in units of liters)
n is the number of moles
R is the ideal gas constant (R = 8.31 kPa L / (K mol) or R = 0.0821 atm L / (K mol))
T is the temperature (in Kelvin)
First, let's convert the temperature from -23.0 °C to Kelvin:
T = -23.0 + 273.15 = 250.15 K
Now we can plug in the values into the ideal gas law equation:
P * 8.90 L = 0.150 mol * 8.31 kPa L / (K mol) * 250.15 K
Simplifying the equation:
P = (0.150 mol * 8.31 kPa L / (K mol) * 250.15 K) / 8.90 L
After canceling out units and performing the calculation, you should find the value of P in kPa.
Keep in mind that the units need to be consistent throughout the equation. In this case, we are using Kelvin for temperature, liters for volume, and kilopascals for pressure.