A 3-L flask contains nitrogen gas at 21°C and 0.95atm pressure. What is the final pressure in the flask if an additional 4 g of N2 gas is added to the flask and the flask cooled to -59°C?
To find the final pressure in the flask, we can use the ideal gas law equation: PV = nRT, where P is the pressure, V is the volume, n is the number of moles of gas, R is the ideal gas constant, and T is the temperature.
First, let's calculate the number of moles of nitrogen gas in the 3-L flask at 21°C and 0.95 atm pressure. We can use the following formula to calculate the number of moles (n):
n = PV / RT
Given:
P1 = 0.95 atm (initial pressure)
V1 = 3 L (initial volume)
T1 = 21°C + 273.15 = 294.15 K (initial temperature)
R = 0.0821 L·atm/(mol·K) (ideal gas constant)
n1 = (0.95 atm * 3 L) / ( 0.0821 L·atm/(mol·K) * 294.15 K)
n1 ≈ 0.11 mol
Now, let's calculate the number of moles of the additional 4 g of N2 gas that is added to the flask. We can use the formula:
n2 = m / M
Given:
m = 4 g (mass of N2 gas)
M = 28.0134 g/mol (molar mass of N2 gas)
n2 = 4 g / 28.0134 g/mol
n2 ≈ 0.143 mol
To find the total moles (ntotal), we add n1 and n2:
ntotal = n1 + n2
ntotal ≈ 0.11 mol + 0.143 mol ≈ 0.253 mol
Now, let's find the final pressure by rearranging the ideal gas law equation:
P2 = (ntotal * R * T2) / V
Given:
T2 = -59°C + 273.15 = 214.15 K (final temperature)
V = 3 L (volume)
P2 = (0.253 mol * 0.0821 L·atm/(mol·K) * 214.15 K) / 3 L
P2 ≈ 0.465 atm
Therefore, the final pressure in the flask after adding 4 g of N2 gas and cooling to -59°C is approximately 0.465 atm.