IB · B · structured
The particulate nature of matter · Question 10
The particulate nature of matter · Original GioPhysics question with a detailed, mark-by-mark answer guide.
- Demand
- discriminating
- Marks
- 13
- Topics
- 2
- Answer
- Complete
Part 1. A cell of e.m.f. 6.0 V and internal resistance 0.75 Ω is connected to an external resistor of resistance 3.0 Ω. Part 2. A fixed mass of ideal monatomic gas is taken from state X to state Y by two different routes: an isothermal expansion, and an adiabatic expansion to the same final volume.
- (a)
Calculate Calculate the current in the circuit and the terminal potential difference of the cell.
3 marks - (b)
Determine Determine the percentage of the power produced by the cell that is dissipated inside the cell itself.
3 marks - (c)
Compare Compare the final temperature and the work done by the gas for the isothermal and the adiabatic expansions to the same final volume.
4 marks - (d)
Explain Explain why the entropy of the gas increases during the isothermal expansion but is unchanged during a reversible adiabatic expansion.
3 marks
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(a)
Calculate Calculate the current in the circuit and the terminal potential difference of the cell.
List the given quantities with units, identify the required quantity, write the governing relationship before substituting, and keep extra digits until the final line so rounding does not distort the result.
- 1
I = ε/(R + r) = 6.0/3.75 = 1.6 A
- 2
V = IR = 1.6 × 3.0
- 3
V = 4.8 V
(b)
Determine Determine the percentage of the power produced by the cell that is dissipated inside the cell itself.
List the given quantities with units, identify the required quantity, write the governing relationship before substituting, and keep extra digits until the final line so rounding does not distort the result.
- 1
power in r = I²r = 1.6² × 0.75 = 1.92 W
- 2
total power = εI = 6.0 × 1.6 = 9.6 W
- 3
percentage = 1.92/9.6 = 20%
(c)
Compare Compare the final temperature and the work done by the gas for the isothermal and the adiabatic expansions to the same final volume.
List the given quantities with units, identify the required quantity, write the governing relationship before substituting, and keep extra digits until the final line so rounding does not distort the result.
- 1
isothermal: the temperature is unchanged by definition, so ΔU = 0 and the work done by the gas equals the thermal energy absorbed
- 2
adiabatic: no thermal energy enters, so the work done by the gas comes entirely from its internal energy
- 3
the adiabatic expansion therefore ends at a lower temperature than the isothermal one
- 4
and because its pressure falls faster with volume, the area under its curve is smaller — less work is done by the gas
(d)
Explain Explain why the entropy of the gas increases during the isothermal expansion but is unchanged during a reversible adiabatic expansion.
List the given quantities with units, identify the required quantity, write the governing relationship before substituting, and keep extra digits until the final line so rounding does not distort the result.
- 1
entropy change for a reversible process is ΔS = Q/T
- 2
in the isothermal expansion the gas absorbs thermal energy at constant temperature, so Q is positive and ΔS is positive — and in molecular terms the same molecules now occupy a larger volume, so there are more available microstates
- 3
in a reversible adiabatic expansion Q = 0, so ΔS = 0: the increase in volume is exactly offset by the narrowing of the molecular speed distribution as the gas cools
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