Two ideal gases in separate rigid containers have pressures 100 kPa and 200 kPa but both have temperature 300 K. Which statement about thermal equilibrium is correct?
(a)They are in thermal equilibrium despite their unequal pressures.
(b)They cannot be in thermal equilibrium unless their masses match.
(c)They cannot be in thermal equilibrium until their pressures match.
Show answer
(a) They are in thermal equilibrium despite their unequal pressures.
Thermal equilibrium requires equal temperature. Mechanical constraints can maintain different pressures in the two rigid containers.
First law, signs and path independence
Question 2
A gas goes from A to B, absorbing 150 J and doing 100 J of work. A second path joins the same states and absorbs 130 J. How much work does the gas do on the second path?
(a)180 J
(b)80 J
(c)It cannot be determined because ΔU can differ between the two paths.
Show answer
(b) 80 J
The first path gives ΔU=Q−W=50 J. The second path has the same ΔU, so its work equals its heat input minus 50 J.
Internal energy and modes of energy transfer
Question 3
An ideal-gas container is smoothly accelerated as a whole so its centre-of-mass speed changes from v to 3v. Its internal temperature and configuration are maintained. What happens to internal energy?
(a)It stays unchanged.
(b)It increases because the bulk kinetic energy increases.
(c)It is multiplied by 3.
Show answer
(a) It stays unchanged.
Internal energy excludes centre-of-mass kinetic energy. Changing bulk speed alone does not change the specified internal state.
Isochoric and isobaric energy balances
Question 4
A gas absorbs 100 J and expands quasi-statically at constant pressure 20 Pa through volume change 2 m³. What is ΔU?
(a)100 J
(b)140 J
(c)60 J
Show answer
(c) 60 J
The gas does work PΔV. Subtract this work from the supplied heat: ΔU=Q−PΔV.
Reversibility requires restoration of system and surroundings
Question 5
Consider a piston moving extremely slowly against dry friction. Does making its motion slower by itself make the process reversible?
(a)No; dissipation prevents exact reversal of system and surroundings.
(b)Yes; friction and viscosity disappear merely because motion is slow.
(c)Yes; every sufficiently slow process is reversible.
Show answer
(a) No; dissipation prevents exact reversal of system and surroundings.
Quasi-static motion is a necessary idealisation but does not remove dissipation by itself. Reversibility also requires no frictional or viscous loss.
Heat capacity, calorimetry and heating rates
Question 6
Two samples of the same substance at the same temperature have masses m and 2m. They undergo the same process without a phase change. What is smaller-sample heat capacity/larger-sample heat capacity?
(a)1/2
(b)1
(c)2
Show answer
(a) 1/2
Total heat capacity S=ms. Specific heat s is unchanged, so total capacity scales with mass.
Ideal-gas heat capacities and Cp−Cv
Question 7
Room-temperature ideal diatomic nitrogen of mass 28 g is heated by 20 K at constant pressure. Take molar mass 28 g/mol, R=8.3 J/(mol K), Cv=5R/2. What heat is required, rounded to two decimal places where needed?
(a)415 J
(b)166 J
(c)581 J
Show answer
(c) 581 J
At constant pressure Cp=Cv+R=7R/2. Use Q=(mass/28)CpΔT, including the expansion-work contribution.
Isothermal ideal-gas energy and work
Question 8
An ideal gas expands reversibly and isothermally from V to 3V with nRT=200 J. What work does the gas do?
(a)200 ln(3) J
(b)400 J
(c)-200 ln(3) J
Show answer
(a) 200 ln(3) J
Pressure follows P=nRT/V throughout the path. Integrating gives W=nRT ln(V2/V1), retaining the logarithm exactly.
State variables, equations of state and scaling
Question 9
A uniform equilibrium gas at 300 K with internal energy 800 J is divided into 2 equal portions without changing their local states. What are temperature and internal energy of one portion?
(a)150 K; 400 J
(b)300 K; 400 J
(c)300 K; 800 J
Show answer
(b) 300 K; 400 J
Temperature is intensive and stays fixed. Internal energy is extensive and divides in proportion to the gas amount.
Quasi-static paths and insulated free expansion
Question 10
A fixed ideal gas initially at 3 atm freely expands into vacuum in an insulated rigid enclosure until its occupied volume is 3 times larger. What is final equilibrium pressure? Assume the heat-capacity ratio γ is greater than 1.
(a)3/3^γ atm
(b)9 atm
(c)1 atm
Show answer
(c) 1 atm
Free expansion gives Q=W=0 and hence unchanged ideal-gas temperature. Compare only equilibrium endpoints: P2=P1/3. The reversible-adiabatic law does not apply to the intermediate free expansion.
Reversible adiabatic relations and temperature change
Question 11
An ideal gas is compressed reversibly and adiabatically to 1/4 of its initial volume. Its constant heat-capacity ratio is γ=3/2. What is final pressure/initial pressure?
(a)1/4
(b)8
(c)4
Show answer
(b) 8
Use P1V1^γ=P2V2^γ. The pressure ratio is the compression ratio raised to γ, not to 1.
Cyclic state changes and signed PV work
Question 12
Using the pressure-volume graph, find total work done by the gas along D→E→F. Work by the gas is positive.
(a)-450 J
(b)450 J
(c)2250 J
Show answer
(b) 450 J
The sloping expansion does work equal to its trapezoidal area. The horizontal compression does negative work; subtract that rectangle to leave the triangular area.
Question 13
The pressure-volume graph shows a cycle traversed D→F→E→D. What is the net heat absorbed by the gas, with heat into it positive?
(a)450 J
(b)2250 J
(c)-450 J
Show answer
(c) -450 J
The vertical leg does no work. The low-pressure expansion is outweighed by the sloping compression. Signed net work is minus the enclosed area; since the cycle has ΔU=0, net absorbed heat has the same negative value.
Second-law restrictions, engines and refrigerators
Question 14
A proposed cyclic device extracts 300 J from one reservoir, delivers 300 J of work and causes no other change. Is this possible?
(a)It is possible because energy is conserved.
(b)It violates energy conservation because heat cannot ever become work.
(c)It conserves energy but violates the Kelvin–Planck statement.
Show answer
(c) It conserves energy but violates the Kelvin–Planck statement.
Its first-law balance closes, but that does not establish feasibility. Complete cyclic conversion of heat from one reservoir with no other effect is forbidden by the second law.
Carnot cycle, temperature limit and working substance
Question 15
During the adiabatic compression from the cold isotherm to the hot isotherm in an ideal Carnot cycle, how much heat is exchanged?
(a)Nonzero heat, because temperature changes during that leg.
(b)Zero, despite the temperature change.
(c)The full heat intake of the hot isotherm.
Show answer
(b) Zero, despite the temperature change.
Adiabatic legs are insulated and have Q=0. Work changes internal energy and temperature during those legs.
Does zero heat transfer mean constant temperature?
Wrong
An insulated gas stays at the same temperature when compressed.
Why it feels right. I associate temperature change with heating only.
Right
Work can change internal energy even when heat transfer is zero.
Worked example
Insulated compression
Heat transfer ≈ 0→Piston does work on gas→Internal energy and ideal-gas temperature rise
Insulation stops heat crossing the boundary. The moving piston still transfers energy by work.
Compress a gas in an insulated cylinder.
Work done on the gas raises its internal energy.
For an ideal gas this raises temperature.
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