Question

What is the volume occupied by 35.4 grams of nitrogen gas at 35 degrees celsius and 735 torr?

Answer 1

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 2

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