IB Physics SL B.3 Gas laws questions

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Type
Multiple choiceWritten
Difficulty
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Question 1

Paper 2Medium8 marks

An ideal gas is in a vessel of uniform pressure. Figure 1 shows the gas and a surface SS of the vessel. The force shown is the resultant force exerted by the gas on SS

Use NA=6.02×1023 mol−1N_A=6.02\times10^{23}\ \mathrm{mol^{-1}} and R=8.31 J mol−1 K−1R=8.31\ \mathrm{J\,mol^{-1}\,K^{-1}} Give answers to three significant figures, except in (d), where give the temperature to the nearest degree Celsius.

surface S, A = 2.40 × 10−3 m2force exerted by gas, F = 240 NN = 2.50 × 1023 moleculesV = 1.00 × 10−2 m3
(a)

Determine the pressure of the gas.

[2]
(b)

Determine the amount of substance of the gas.

[2]
(c)

Using your answers to (a) and (b), determine the temperature of the gas in kelvin.

[3]
(d)

Express the temperature from (c) in degrees Celsius.

[1]

Question 2

Multiple choiceHard1 mark

A sample of 6.0×10236.0 \times 10^{23} particles of a monatomic gas is initially at 400 K400\ \mathrm{K} In a final state, its volume is twice its initial volume, its number density is 3/43/4 of the initial value, and its temperature is 2/32/3 of the initial value. At both states the gas is sufficiently hot and dilute for the ideal-gas approximation to apply. What is its final internal energy? Take kB=1.38×10−23 J K−1k_B=1.38 \times 10^{-23}\ \mathrm{J\,K^{-1}}

Show answer and explanation

Answer: C. The particle-number ratio is the number-density ratio multiplied by the volume ratio: Nf/Ni=(3/4)(2)=3/2N_f/N_i=(3/4)(2)=3/2 Since U=32NkBTU=\frac{3}{2}Nk_BT the internal-energy ratio is (3/2)(2/3)=1(3/2)(2/3)=1 Thus Uf=Ui=32(6.0×1023)(1.38×10−23)(400)=4.968×103 JU_f=U_i=\frac{3}{2}(6.0\times10^{23})(1.38\times10^{-23})(400)=4.968\times10^3\ \mathrm{J} or 5.0×103 J5.0\times10^3\ \mathrm{J} to two significant figures.
A: treats the number-density ratio as the particle-number ratio, omitting the volume change.
B: accounts for the temperature change but holds the particle number fixed.
D: treats the volume ratio as the particle-number ratio, ignoring the number-density change.
Remember: for an ideal monatomic gas, internal energy depends on particle number and absolute temperature.

Question 3

Paper 2Hard10 marks

A fixed amount of gas is measured in a cylinder with a movable piston. Figure 1 shows the measured pressure–volume states of the gas at two temperatures. The temperatures are in kelvin. XX is a further state of the same gas, at an unrecorded temperature. No curves have been fitted to the measurements.

00.511.522.533.5020406080100120140160180200V / LP / kPa● 300 K■ 450 K◆ X◆
(a)

Describe the pressure–volume relationship indicated by the 300 K300\ \mathrm{K} measurements.

[1]
(b)

For the measurements at V=2.00 LV=2.00\ \mathrm{L} calculate PV/TPV/T at each temperature. Give an appropriate unit.

[2]
(c)

A fixed amount of gas changes from state 1 (P1,V1,T1)(P_1,V_1,T_1) to state 2 (P2,V2,T2)(P_2,V_2,T_2) Using the empirical pressure–temperature law at constant volume and Boyle’s law at constant temperature, derive a relation between these two states. Temperatures are in kelvin.

[2]
(d)

A controller is expected to switch when the gas reaches 750 K750\ \mathrm{K} Use state XX and both calibration results from part (b) to estimate the temperature at XX and decide whether the controller’s reported temperature is consistent with the measurements. Show your working.

[5]

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