Q#1 (Past Exam Paper – June 2011 Paper 42 & 43 Q2)
(a) State what is meant by a mole. [2]
(b) Two containers A and B are joined by a tube of negligible volume, as illustrated in
Fig. 2.1.
Fig. 2.1
The containers are filled with an ideal gas at a pressure of 2.3 × 105 Pa.
The gas in container A has volume 3.1 × 103 cm3 and is at a temperature of 17 °C.
The gas in container B has volume 4.6 × 103 cm3 and is at a temperature of 30 °C.
Calculate the total amount of gas, in mol, in the containers. [4]
Solution:
(a) A mole is an amount of a substance containing the same number of particles (atoms / molecules) as in 0.012kg of carbon-12.
(b)
pV = nRT
(n = pV / RT)
{total amount of gas, in mol, n = nA + nB}
Total amount = (2.3×105 × 3.1×10-3) / (8.31×290) + (2.3×105 × 4.6×10-3) / (8.31×303)
Total amount = 0.296 + 0.420 = 0.716 mol
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