Concepts Tests
1-0 Which has higher internal energy?
Gas cylinder, 2000 psi (KE only) or
Snickers bar + oxygen, 280 Cal (chemical only)
Green: Snickers has way more Pink: Cylinder has way more
Blue: Comparable U
Answer: Comparable. In class, we worked out that the internal energy [U = (3/2)NkT =
(3/2)pV] for a tank of He gas at 2000 psi is about 106 J.
Snickers bar has 280 kcal  4.2 J/cal  1.2 x 106 J.
How can a little Snicker’s bar have as much stored energy as that big tank of compressed
He? The energy per atom of the He is kTroom = 1/40 eV. The energy of a typical
chemical bond is roughly 1 eV. So the energy per atom in Snickers bar is roughly 1 eV.
1-1
A gas at temperature T is mixture of hydrogen and helium gas.
Which particles have more KE (on average)?
Green: H2
Pink: He
Blue: both have same KE
Answer: Both have the same average KE = (3/2)kT, by the Equipartition Partition. The
hydrogen molecule has more total energy but the same average KE. He gas has 3
degrees of freedom per atom. The H2 molecule can rotate (around 2 independent axes) as
well as translate, so it has 5 degrees of freedom. At room temp, the vibrational degrees of
freedom are “frozen out”.
1-2
How many degrees of freedom in a 1D simple harmonic oscillator?
Green: 1
Pink: 4
Blue: Some other number
Answer: f = 2 This is a 1D oscillator; only motion along the x-direction is allowed. Etot
= KE + PE = (1/2)mv2 + (1/2)kx2 There are two quadratic terms in the expression for
energy, so f = 2.
1-3
A resistor is heated by an electrical current. Is this energy transfer an example of heat or
work?
Green: Heat
Pink: Work
Blue: Neither
Answer: Work. Heat transfer occurs only if energy flows from a high temperature object
to a low temperature object. In this case, the energy is transformed from electrostatic
potential energy of the conduction electrons into kinetic/vibrational energy of the atoms
in the resistor. The conduction electrons and atoms are both at the same temperature.
1-4 I lift a rock a upward a height h. The work done by the force of gravity is
Green: -mgh Pink: +mgh
Blue: depends on how I lifted the rock
Answer: -mgh displacement is upward, force of gravity is downward, so W = force dot
displacement is negative.
1-5
A gas expands according to this pV diagram. The work done by the gas system on the
environment is Green: positive
Pink: negative
Blue: zero
Answer: Work done by the gas is positive (force and displacement in same direction).
Work done on the gas by the environment is negative (force and displacement in opposite
directions).
V
p
p2
p
p1
V
V1
V2
True (Green) or False(Pink):
The magnitude of the work done is
p 1D V + D p D V
Answer: False. The area under the curve is
p 1D V + (1/ 2)D p D V
1-7 An ideal gas expands at constant pressure. Its temperature T must..
Green: increase
Pink: decrease
Blue: can’t tell
Answer: T must increase, since the product pV increased and pV = NkT.
p
p1
V
V1
V2
Its internal energy U must..
Green: increase
Pink: decrease
Blue: can’t tell
Answer: U must increase. Since T increased, and U = (f/2)NkT
Heat Q must have been ..
Green: added (Q+) Pink: removed (Q-) Blue: can’t tell
Answer: Not trivial! 1st Law of Thermodynamics: U = W+Q. In this case, U is (+)
and W is (-). (Work done on gas is negative of area under curve.) So U = W+Q implies
(+) = (-) + ? The ? must be positive. (At constant pressure, the way to expand a gas is to
add heat.)
A gas is taken through the complete cycle shown.
p
1.8 The change in internal energy U must be
Green: positive
Pink: negative
Blue: zero
Answer: U = 0. The system has returned to its initial starting
state. Internal energy is a “function of state”, meaning U has a
single-value, depending only on the state of the system (location
in pV plane).
Alternative explanation: If the gas returns to the same (p, V)
point, then it has returned to the same temperature T (since pV
= NkT). Since U = N(f/2)kT, then no change in T means no
change in U.
Start/finish
V
p
p
W+
WV
V
The net work done on the system was
Green: positive W+ Pink: negative WBlue: zero
Answer: Positive. Add up negative work done in first leg of loop and positive work done
in 3rd leg. (No work done in legs 2 and 3).
Sign of net heat added to the system during the cycle?
Green: positive Q+ Pink: negative QBlue: zero
Answer: Q must be negative. U = 0 = W + Q. W is positive so Q must be negative.
Net heat has extracted from the gas.