What is the volume occupied by 35.4 grams of nitrogen gas at 35 degrees celsius and 735 torr?
735 torr = 1 atmosphere(atm)
Temperature in Kelvin = 273 + 35 = 308 K
Moles of Nitrogen gas = mass of N2 / molar mass of N2
= 35.4/28.014
= 1.26 moles
Using Ideal gas equation,
PV = nRT
1 x V = 1.26 x 0.082 x 308
V = 31.822 liters.
Hope this helps!
Answer : The volume occupied by the nitrogen gas is, 33.06 L
Explanation :
Using ideal gas equation,
\(PV=nRT\)
\(PV=\frac{w}{M}RT\)
where,
P = pressure of nitrogen gas = 735 torr = 0.967 atm
conversion used : (1 atm = 760 torr)
V = volume of nitrogen gas = ?
T = temperature of nitrogen gas = \(35^oC=273+35=308K\)
n = number of moles of nitrogen gas
R = gas constant = 0.0821 L.atm/mole.K
w = mass of nitrogen gas = 35.4 g
M = molar mass of nitrogen gas = 28 g/mole
Now put all the given values in the ideal gas equation, we get the volume of nitrogen gas.
\((0.967atm)\times V=\frac{35.4g}{28g/mole}\times (0.0821L.atm/mole.K)\times (308K)\)
\(V=33.06L\)
Therefore, the volume occupied by the nitrogen gas is, 33.06 L