At what pressure would 0.150 mole of nitrogen gas at -23.0^o C occupy 8.90 L ?
P=nRT/V
change temp to kelvins
thanks bob
P = nRT / V
P = [ (0.150 mol) (0.08206 L atm mol¯1 K¯1) (296.0 K) ] / 8.90 L
To find the pressure of the nitrogen gas, we can use the ideal gas equation:
PV = nRT
Where:
P = pressure
V = volume
n = number of moles of gas
R = gas constant (0.0821 L·atm/(mol·K))
T = temperature in Kelvin
First, we need to convert the temperature from Celsius to Kelvin:
T(K) = T(°C) + 273.15
So, -23.0°C + 273.15 = 250.15 K
Now, we can substitute the values into the equation:
PV = nRT
P * 8.90 L = 0.150 mol * 0.0821 L·atm/(mol·K) * 250.15 K
Simplifying the equation:
P * 8.90 = 0.150 * 0.0821 * 250.15
P * 8.90 = 3.02286675
P = 3.02286675 / 8.90
P = 0.3396 atm
Therefore, at a temperature of -23.0°C, nitrogen gas would occupy a pressure of approximately 0.3396 atm.