A lvl H2 Phy: Thermal Physics/Ideal Gas

A sealed container holds a mixture of nitrogen molecules and helium molecules at a temperature of 290K. The total pressure exerted by the gas on the container is 120 kPa.

(M) molar mass of helium = 4 x 10^-3 kg / mol
R= 8.31 J / K mol
Na= 6.02 x 10²³ /mol

i) calculate root mean square speed of the helium molecules.

ii) calculate average kinetic energy of a nitrogen molecule.

iii)If there are twice as many helium molecules as nitrogen molecules in the container, calculate pressure exerted on the container by helium molecules.

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Answer:


i) Average kinetic energy per mole = (1/2) * Na*m v² = 3/2 RT
Since M = Na * m= 4 x 10^-3 kg / mol,
then, v² = ( (3 * R T) / M = 3 * 8.31 * 290 / (4 x 10^-3) =1.81 x 10⁶
v_rms = √(1.81 x 10⁶)
= 1346 m/s

ii) Average kinetic energy of a nitrogen molecule
= 3/2 KT, where K = boltzmann constant
= 3/2 * 1.38 * 10^-23 * 290
= 6.00 *10^-21 J

iii) pV = nRT
From here, we can see that R is constant, T and V is same for both helium and nitrogen. The only difference is n.
So, we can expect that helium molecules will exert twice the amount of pressure as the nitrogen molecules in this question

Hence, pressure exerted on the container by helium molecules is 80 kPa.

A lvl H2 Phy: Thermal Physics/Ideal Gas

State, in words, the relation between the increase in the internal energy of a gas, the work done on the gas, and the heat supplied to the gas.

(a) A quantity of 0.200 mol of air enters a diesel engine at a pressure of 1.04 * 10⁵ Pa and at a temperature of 297 K. Assuming that air behaves as an ideal gas, find the volume of this quantity of air.

(b) The air is then compressed to one twentieth of this volume, the pressure having risen to 6.89 * 10⁵ Pa. Find the new temperature.

(c) Heating of the air then takes place by burning a small quantity of fuel in it to supply 6150 J. This is done at a constant pressure of 6.89 * 10⁵ Pa as the volume of air increases and the temperature rises to 2040 K. Find
(i) the volume of the gas after burning the fuel,
(ii) the work done by the air during this expansion,
(iii) the change in the internal energy of the air during this expansion.

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Answer:

The increase in internal energy of a gas is equal to the sum of the work done on the gas and the heat supplied to the gas.

(a) PV = nRT
V = nRT/P
= (0.200 * 8.31 * 297) / (1.04 * 10⁵)
= 4.75 * 10^-3 m³


(b) Given V' = V/20, and P' = 6.89 * 10⁵ Pa,




= 98.4 K


(c)
(i) T₁/T₂ = V₁ / V₂ (when pressure is constant)
V₂ = 2040 / 98.4 * V'
V₂ = 2040 / 98.4 * 2.37 * 10^-4
= 4.92 * 10^-3 m³

(ii) Work done by air
= p ΔV
= 6.89 * 10⁵ * (4.92 - 0.237) * 10^-3
= 3230 J

(iii) ΔU = Q supplied + WD on air
= Q supplied - WD by air
= 6150 - 3230
= 2920 J