Many analytical instruments require liquid nitrogen. While filling an instrument with liquid nitrogen, an analytical chemist slips and spills the contents of the dewer, 25L, onto the floor. Since the boiling point of liquid nitrogen at 1atm is 77.4 K, all the nitrogen boils away to gas very quickly. The density of liquid nitrogen is 804kg/m^3. What volume of nitrogen gas is produced from the liquid nitrogen spill if the temperature of the room is 25 degrees C at 1 atm?
Is my calculations correct? Do i calculate the 77.4K in the problem?
If density = 804kg/m³
Mass 1 litre = 804g
25l = 25*804 20,125g
Molar mass N2 = 28g/mol
Mol of N2 = 20,125/28= 718.75 mol
At STP 1mol = 22.4 litres
Volume at STP = 718.75*22.4 = 16,100 litres
This is at 273K and 1 atm pressure
At 25°C = 298K and 1 atm
V1/T1 =V2/T2 (P is constant at 1 atm)
V1/298 = 16100/273
V1 = 16100*298/273
V1 = 17,574 litres.
Volume of N2 gas produced = 17,574 litres.
Is this right?????
25*804 = 20,100 and that error carries through. I think it's easier to convert 20,100 to moles and use PV = nRT but I don't see anything wrong with yours other than the initial multiplication error.
Your calculations are correct, but there is a small mistake in the conversion from grams to kilograms. The mass of 25 liters of liquid nitrogen is 20,125 grams, which is equal to 20.125 kilograms, not 20,125 kilograms. Other than that, your calculations are accurate.
To answer your question about whether to calculate the boiling point of liquid nitrogen at 77.4 K, the answer is no. The given boiling point of 77.4 K is only provided to give you the information that all the liquid nitrogen boils away quickly when spilled. In this case, you are not required to calculate the exact boiling point temperature. Instead, you only need to use the given room temperature of 25 degrees Celsius (298 K) to determine the volume of nitrogen gas produced.