Question

At what temperature will 41.6 grams N, exert a pressure of 815 mmHg in a
20.0L cylinder?
Hint: You must convert 41.6 g into moles FIRST, then use Ideal gas law to solve
for temperature.
134K
O 238 K
O 337 K.
O 176K

Answer 1

None of the given options are correct, and the correct answer is 174 K.

Explanation:

Mass of N = 41.6 g

Moles = $\frac{given mass}{molar mass}$

      = $\frac{41.6 g}{28 g/ mol}$ = 1.5 moles

Pressure, P = 815 mm Hg = 1.07 atm

Volume, V = 20 L

R = gas constant = 0.08205 L atm mol⁻¹ K⁻¹

Temperature, T = ? K

We have to use the ideal gas equation,  

PV = nRT  

by rearranging the equation, so that the equation becomes,  

T = $\frac{PV}{nR}$

 Plugin the above values, we will get,  

T = $\frac{1.07 \times 20}{1.5 \times 0.08205}$

  = 174 K

So the temperature of the nitrogen gas is 174 K.