A given amount of gas occupies 10.0dm\(^3\) at 4 atm and 273°C. The number of moles of the gas present is?
[Molar volume of gas at s.t.p = 22.4 dm\(^3\)]
A
0.89 mol
B
1.90 mol
C
3.80 mol
D
5.70 mol
Verified Correct Answer: A.
Step-by-Step Mathematical Solution
Given:
P = 4 atm , V = 10 dm\(^3\) , n = ? R = 0.082 (constant) T = 273 + 273 = 546K
Using the ideal gas law, PV = nRT
n = \(\frac{PV}{RT}\)
n = \(\frac{{4}\times{10}}{{0.082}\times{546}}\)
n = \(\frac{40}{44.772}\)
n = 0.893 mole
Source: WAEC / JAMB / NECO Classroom Archive (Record #47284)