3 free questions from this chapter (the full set has 30). Marking: Multi-correct: +4 all correct, +1 per correct option if none wrong, −2 if any wrong · Numerical: +4 / 0 · Matching: +3 / −1.
Question 1Level 2 · ExamSingle correct
Three rigid vessels of volumes V, 2V and 3V are joined by thin tubes of negligible volume, and together contain an ideal gas at a uniform pressure P0 and temperature 300 K. The vessels are now kept at different temperatures: the vessel of volume V at 300 K, the vessel of volume 2V at 400 K and the vessel of volume 3V at 600 K. The tubes stay open; neglect heat conduction along them and the thermal expansion of the vessels. In the final steady state the common pressure of the gas is ______. Exactly one option is correct.
A34P0
B1829P0
C23P0
D59P0
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Answer: C
The total amount of gas is fixed and in the steady state the pressure P is the same everywhere, while each vessel has its own temperature.
n=R(300)P0(6V)=RP(300V+4002V+6003V)=RPV⋅6008.
So 60012P0=6008P, giving P=23P0.
Trap: 1829P0 comes from using the volume-weighted mean temperature (6300+800+1800=483.3 K) with PV=nRT; it is the reciprocal temperatures that add, because the amount of gas per unit volume is P/(RT).
Difficulty: HardQuestion 2Level 2 · ExamSingle correct
A hot-air balloon has a fixed envelope volume of 400 m3 and is open at the bottom, so the air inside is always at the pressure of the surrounding air. The outside air is at 300 K, where its density is 1.20 kg/m3. The total mass of the envelope fabric, basket and load (excluding the air inside the envelope) is 120 kg. Treat the air inside and outside as the same ideal gas, neglect the volume of everything except the envelope, and neglect any variation of pressure and temperature with height across the balloon. The minimum temperature of the air inside the envelope for the balloon to just lift off is ______. Exactly one option is correct.
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Answer: A
At equal pressure, ρ∝1/T for the same gas, so the inside density is ρin=ρ0TT0.
Lift-off: buoyancy = weight of the hot air + weight of the load: ρ0Vg=ρ0TT0Vg+mg.
ρ0V(1−TT0)=m⇒480(1−T300)=120⇒T300=43⇒T=400 K=127∘C.
Trap: 36∘C is 27∘C/0.75, i.e. Celsius used in the gas law.
Difficulty: HardQuestion 3Level 3 · AdvancedSingle correct
A horizontal cylinder with rigid, thermally insulating walls is closed at both ends and divided into two parts by a frictionless piston that conducts heat but has negligible heat capacity. Initially the piston is held fixed; the left part contains 2 mol of helium at 300 K and the right part contains 1 mol of nitrogen at 600 K. Both gases are ideal; treat nitrogen as a rigid diatomic gas (no vibration). The piston is released and the system is left until it reaches equilibrium. The final temperature of the gases (to the nearest kelvin) is ______. Exactly one option is correct.
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Answer: D
System = both gases. The walls are rigid and insulated, so the system exchanges neither heat nor work with the surroundings. The piston (no mass effect at equilibrium, no heat capacity) only transfers energy from one gas to the other, as heat and as work. Hence the total internal energy is conserved:
n1CV1(T−300)+n2CV2(T−600)=0 with n1CV1=2⋅23R=3R, n2CV2=1⋅25R=2.5R.
T=5.53(300)+2.5(600)=5.52400=436.4 K≈436 K.
(At the end the pressures are equal and the temperatures are equal, so the volumes divide in the ratio 2:1; this does not affect T.)
Traps: 424 K weights by Cp ("each gas is at constant pressure"); 400 K weights by moles only; 450 K is the plain average.
Insight: although the piston moves and each gas does work on the other, the work is internal to the two-gas system, so only ∑nCVT is conserved.
Difficulty: Very hard