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(A) 7.84 
× 10
10
ergs
(B) 1.50 
× 10
11
ergs
(C) 6.00 
× 10
11
ergs
(D) 1.87 
× 10
10
ergs
(E) 1.00 
× 10
12
ergs
The correct answer is (A). To solve this problem one must remember that the total kinetic energy of an
ideal gas depends only on the number of moles of gas present and not the molecular weight of the gas.
Using the formula of kinetic energy = 3/2nRT where n equals the number of moles, one can calculate that
two moles of O
2
have twice the kinetic energy of one mole of He
2
at the same temperature.
EXAMPLE:
A known volume of nitrogen takes 15 seconds to diffuse through a small pinhole under
constant pressure. The same volume of an unknown gas, X, takes 30 seconds to diffuse through the
pinhole under identical conditions. What is the molecular weight of gas X?
(A) 65
(B) 82
(C) 100
(D) 112
(E) 130

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Lesson 10
284
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ARCO
■ SAT II Subject Tests
The correct answer is (D). This problem is an application of the equation:
Using the given information, we have:
EXAMPLE:
Fifteen kcal is added to 100g of ice at –20
°C. After equilibrium is reached, one would pre-
dict: (specific heat of water = 1.0kcal/kg
°C; specific heat of ice = 0.5kcal/kg°C; heat of fusion of water =
80kcal/kg; heat of vaporization of water = 540kcal/kg)
(A) 0.1kg of water at 50
°C
(B) 0.1kg of water at 60
°C
(C) 0.1kg of water at 70
°C
(D) 0.1kg of water vapor at 100
°C