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.

Step-by-step explanation:

Mass of N = 41.6 g

Moles =
`$(given mass)/(molar mass)`

=
`$(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 =
`$(PV)/(nR)`

Plugin the above values, we will get,

T =
`$(1.07 * 20)/(1.5 * 0.08205)`

= 174 K

So the temperature of the nitrogen gas is 174 K.