Find temperature of 20.0g of nitrogen that occupies volume of 500.0ml at pressure 0.45 bar.
PV=nRT
T=PV/nR
To find the temperature of nitrogen, we can use the Ideal Gas Law equation, which states:
PV = nRT
Where:
P is the pressure in atm (1 bar = 1 atm)
V is the volume in liters (1 L = 1000 mL)
n is the number of moles of gas
R is the ideal gas constant (0.0821 L·atm/(mol·K))
T is the temperature in Kelvin
First, we need to convert the given values to the appropriate units:
20.0 g of nitrogen = 20.0 g
500.0 mL = 500.0 mL / 1000 mL/L = 0.500 L
Pressure = 0.45 bar = 0.45 atm
Next, we need to find the number of moles of nitrogen. To do this, we can use the ideal gas equation rearranged to solve for n:
n = PV / RT
Substitute the known values:
n = (0.45 atm) * (0.500 L) / [(0.0821 L·atm/(mol·K)) * T]
Now, we can determine the temperature (T) of nitrogen by rearranging the equation and solving for T:
T = PV / (nR)
Substitute the known values:
T = (0.45 atm) * (0.500 L) / [(20.0 g) * (0.0821 L·atm/(mol·K))]
T = 0.01125 mol / 1.642 L·atm/(mol·K)
T = 6.85 K
Therefore, the temperature of 20.0 g of nitrogen, which occupies a volume of 500.0 mL at a pressure of 0.45 bar, is approximately 6.85 Kelvin.