A sample of ideal monatomic gas is cooled by 50.0°C at constant volume by removing 831 J of energy from it.How many moles of gas are in the sample? (R = 8.31 J/mol ∙ K)
A)2.50 mol
B)1.50 mol
C)1.33 mol
D)1.00 mol

Question

Quizplus · Accepted Answer

First, we need to use the formula Q = nCvΔT, where Q is the energy removed from the gas, n is the number of moles of gas, Cv is the molar heat capacity at constant volume (which is 3/2R for a monatomic gas), and ΔT is the temperature change.

Plugging in the given values, we get:

831 J = n * (3/2 * 8.31 J/mol*K) * (-50.0°C + 273.15) K

Simplifying and solving for n, we get:

n = 1.33 mol

Therefore, the answer is choice C.

A Sample of Ideal Monatomic Gas Is Cooled by 50