the ideal gas law, PV = nRT, tells us that P and T are proportional.
Therefore, if T increases by a factor of 2, then so does P.
- E Neither A nor B can be correct. Using PV = nRT, both containers have the
same V, n is the same, P is the same, and R is a universal constant.
Therefore, T must be the same for both samples. C is also wrong, since R is a
universal constant. The kinetic theory of gases predicts that the rms speed of
the gas molecules in a sample of molar mass M and temperature T is
Hydrogen has a smaller molar mass than does helium, so vrms for
hydrogen must be greater than vrms for helium (because both samples
are at the same T).
- A By convention, work done on the gas sample is designated as negative, so in
the first law of thermodynamics, ∆U = Q − W, we must write W = −320 J.
Therefore, Q = ∆U + W = 560 J + (−320 J) = +240 J. Positive Q denotes
heat in.
- C No work is done during the step from state b to state c because the volume
doesn’t change. Therefore, the work done from a to c is equal to the work
done from a to b. Since the pressure remains constant (this step is isobaric),
we find that W = P∆V = (3.0 × 10^5 Pa)[(25 − 10) × 10−3 m^3 ] = 4,500 J.
- B Since the engine takes in 400 J of energy and produces only 400 J −
300 J = 100 J of useful work, its efficiency is.
- D A is the first law of thermodynamics. Choices (C) and (E) are true, but they
are not equivalent to the second law of thermodynamics. B is false, but if it
read, “The efficiency of a heat engine can never be equal to 100 percent,”
then it would be equivalent to the Second Law. Choice (D) is one of the
several equivalent forms of the second law of thermodynamics.
