A given amount of gas occupies 10.0 dm3 at 4atm and 273°C. The number of moles of the gas present is?
(Molar volume of a gas at s.t.p. = 22.4 dm3)
-
0.89 mol
- 1.90 mol
- 3.80 mol
- 5.70 mol
The correct answer is: A
Explanation
\(V_1\) = \(10dm^3\), \(V_2\) = ?, \(P_1\) = 4atm = 4 x 760 = 3040mmHg, \(P_2\) = 760mmHg \(T_1\) = 273ºc = 273 + 273 = 546K, \(T_2\) = 273K
Using the general gas law = \(\frac{P_1 V_1}{ T_1} = \frac{P_2 V_2}{T_2}\)
\(V_2 = \frac{P_1 V_1 T_2}{P_2 T_1}\)
\(V_2 = \frac{ 3040 \times10 \times273}{760 \times546}\)
\(V_2 = 20dm^{-3}\)
1mole of a gas = \(22. 4dm^{3}\)
x = \(20.0dm^{3}\)
= 0.89mol.