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Questions on ideal gases and the gas laws
Boyle's law, Charles's law and the pressure law each hold one quantity fixed, and two of them extrapolate back to a shared absolute zero. An ideal gas obeys pV proportional to T at every pressure, and its equation of state is pV = nRT for moles and pV = NkT for molecules.
21 original questions · 66 marks · the ideal gases and the gas laws notes · Thermal physics
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State Boyle's law for a fixed mass of gas.
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At constant temperature, the pressure of a fixed mass of gas is inversely proportional to its volume (1), so pV = constant (1).A sealed container of volume 0.050 m3 holds 2.0 mol of an ideal gas at 300 K. Calculate the pressure of the gas (R = 8.31 J mol−1 K−1).
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p = nRT/V = (2.0 × 8.31 × 300)/0.050 (1)
p = 9.97 × 104 Pa (1)Explain how extrapolating a graph of pressure against temperature leads to the idea of absolute zero.
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For a fixed volume of gas, pressure falls linearly as temperature falls (1). Extending the straight line back to zero pressure gives the temperature at which the molecules' kinetic energy, and so the gas's internal energy, would be at its minimum, −273°C, which is absolute zero, 0 K (1).A thermometer carried by a weather balloon reads −63°C at high altitude. State this temperature in kelvin.
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T = −63 + 273 = 210 K (1).A rigid sealed flask contains air at a pressure of 1.01 × 105 Pa and a temperature of 20°C. The flask is placed in a water bath at 77°C. Calculate the new pressure of the air.
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Temperatures in kelvin: 293 K and 350 K; at constant volume p/T is constant (1)
p2 = p1T2/T1 = 1.01 × 105 × 350/293 = 1.21 × 105 Pa (1)The gas laws are described as empirical. State what this means.
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They are generalisations found from experiment and observation, not derived from a theory or model (1).A piston slowly compresses the air in a cylinder from 0.020 m3 to 0.005 m3 at constant temperature. The initial pressure is 1.0 × 105 Pa. Calculate the final pressure.
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p1V1 = p2V2 (1)
p2 = (1.0 × 105 × 0.020)/0.005 (1)
p2 = 4.0 × 105 Pa (1)Calculate the volume occupied by 0.50 mol of an ideal gas at a pressure of 2.0 × 105 Pa and a temperature of 300 K, and state why the temperature must be converted to kelvin in this calculation.
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V = nRT/p (1)
V = (0.50 × 8.31 × 300)/(2.0 × 105) (1)
V = 6.23 × 10−3 m3 (1)
The equation needs absolute temperature: only the kelvin scale starts from the true zero of molecular kinetic energy, so ratios of temperatures mean nothing in °C (1)A container of volume 0.010 m3 holds 3.0 × 1023 gas molecules at 290 K. Calculate the pressure using pV = NkT (k = 1.38 × 10−23 J K−1).
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p = NkT/V (1)
p = (3.0 × 1023 × 1.38 × 10−23 × 290)/0.010 (1)
p = 1.20 × 105 Pa (1)A gas of volume 0.010 m3 at 300 K is heated at constant pressure to 400 K. Calculate its new volume.
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At constant pressure V/T is constant (1)
V2 = V1T2/T1 = 0.010 × 400/300 (1)
V2 = 0.0133 m3 (1)A weather balloon is filled with 12.0 m3 of helium at ground level, where the pressure is 101 kPa and the temperature is 288 K. It rises to an altitude where the pressure is 22.0 kPa and the temperature is 218 K. Calculate the volume of the balloon at this altitude.
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For a fixed mass of gas, p1V1/T1 = p2V2/T2 (1)
V2 = 12.0 × (101/22.0) × (218/288) (1)
V2 = 41.7 m3 (1)A cylinder of volume 0.050 m3 contains 0.72 kg of oxygen at 290 K. The molar mass of oxygen is 0.032 kg mol−1 (R = 8.31 J mol−1 K−1). Calculate the pressure in the cylinder.
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n = mass/molar mass = 0.72/0.032 = 22.5 mol (1)
p = nRT/V = (22.5 × 8.31 × 290)/0.050 (1)
p = 1.08 × 106 Pa (1)A student investigating Boyle's law traps a fixed mass of air at constant temperature and records the volume V at four pressures p. p/kPa: 100, 125, 160, 200. V/cm3: 40.0, 32.1, 25.0, 20.1. Deduce whether the results are consistent with Boyle's law.
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Test whether pV is constant (1)
pV/(kPa cm3) = 4000, 4012.5, 4000, 4020 (1)
The products agree to well within 1%, so the results are consistent with Boyle's law (1)A gas expands from 0.010 m3 to 0.015 m3 at a constant pressure of 2.0 × 105 Pa. Calculate the work done by the gas, and state where this energy comes from.
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W = pΔV (1)
W = 2.0 × 105 × (0.015 − 0.010) (1)
W = 1000 J (1)
At constant pressure the temperature rises as the gas expands (pV = nRT), so its internal energy rises too; heating from the surroundings supplies both this and the work done (1)A gas occupies 0.024 m3 at a pressure of 1.0 × 105 Pa and a temperature of 300 K. Calculate the number of molecules present and the corresponding amount of gas in moles (NA = 6.02 × 1023 mol−1).
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From pV = NkT: N = pV/(kT) (1)
N = (1.0 × 105 × 0.024)/(1.38 × 10−23 × 300) = 5.80 × 1023 molecules (1)
n = N/NA = 5.80 × 1023/(6.02 × 1023) (1)
n = 0.96 mol (1)State the conditions under which a real gas behaves most like an ideal gas.
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At low pressure and high temperature, and so at low density, the molecules are far apart (1). Then intermolecular forces are negligible (1) and the molecules' own volume is tiny compared with the container, as the ideal-gas model assumes (1).A party company must fill 250 balloons with helium. Each filled balloon holds 0.012 m3 at a pressure of 1.05 × 105 Pa and a temperature of 293 K. The supplier offers a cylinder of internal volume 0.040 m3 containing helium at 1.5 × 107 Pa, also at 293 K. Helium stops flowing out of the cylinder once its pressure falls to the balloon pressure (R = 8.31 J mol−1 K−1). Deduce whether one cylinder is enough to fill all the balloons.
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Amount per balloon n = pV/RT = (1.05 × 105 × 0.012)/(8.31 × 293) = 0.517 mol (1)
Amount needed = 250 × 0.517 = 129 mol (1)
Amount in cylinder = (1.5 × 107 × 0.040)/(8.31 × 293) = 246.4 mol (1)
Helium left in the cylinder at 1.05 × 105 Pa = 1.7 mol, so the usable amount is 245 mol (1)
245 mol is well above the 129 mol needed, so one cylinder is enough (1)A rigid vessel of volume 0.080 m3 holds nitrogen at a pressure of 1.4 × 105 Pa and a temperature of 300 K. The molar mass of nitrogen is 0.028 kg mol−1 (R = 8.31 J mol−1 K−1, NA = 6.02 × 1023 mol−1). Calculate the density of the nitrogen and the mass of one nitrogen molecule.
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n = pV/RT = (1.4 × 105 × 0.080)/(8.31 × 300) = 4.49 mol (1)
Mass of gas = 4.49 × 0.028 = 0.126 kg (1)
ρ = m/V = 0.126/0.080 = 1.57 kg m−3 (1)
Mass of one molecule = 0.028/(6.02 × 1023) = 4.65 × 10−26 kg (1)Estimate the number of air molecules in a school laboratory at a pressure of 1.0 × 105 Pa and a temperature of 300 K (k = 1.38 × 10−23 J K−1). State the estimate you make for the volume of the room.
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Sensible room volume, for example 7 m × 7 m × 3 m ≈ 147 m3 (accept 50 to 500 m3) (1)
N = pV/kT = (1.0 × 105 × 147)/(1.38 × 10−23 × 300) (1)
N ≈ 3.6 × 1027 molecules (accept order 1027) (1)A student has a rigid flask of air fitted with a pressure sensor, a large water bath, a heater, a thermometer and ice. Describe how the student can use this apparatus to determine a value for absolute zero in °C. Include the measurements taken, the graph to be drawn and how the graph gives the result.
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Submerge the flask in the water bath so the trapped air is at the bath temperature, keeping the volume (and mass) of air constant (1). Record the pressure and the temperature; wait at each temperature for the reading to settle so the air reaches the bath temperature (1). Repeat over the widest possible range of temperatures, using ice and the heater to set them (1). Plot pressure against temperature in °C, which gives a straight line (1). Extrapolate the line back to zero pressure (1). The intercept on the temperature axis is absolute zero, expected near −273°C (1).A fixed mass of gas is sealed in a rigid container of constant volume, and its pressure is measured at five temperatures. Determine, from the readings alone, the temperature at which the pressure of this gas would fall to zero, and state what that temperature is called and why it cannot be reached by extending the measurements.
temperature θ / °C pressure p / kPa 20 107.3 40 114.7 60 122.0 80 129.3 100 136.6 Mark scheme
The readings lie on a straight line, so a gradient can be taken between the extreme points: (136.6 − 107.3)/(100 − 20) (1)
Gradient = 0.366 kPa °C−1 (1)
Working back to 0 °C, p = 107.3 − 0.366 × 20 = 100.0 kPa (1)
p = 0 when θ = −100.0/0.366 = −273 °C (1)
That temperature is absolute zero, 0 K; the line has to be extrapolated far beyond the readings, because the gas would liquefy long before the pressure reached zero (1)
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