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Practice each question, then open the answer or worked solution.
A gas mixture consists of molecules of tyres A, B, and C with masses mA > mB > mC. Rank the three types of molecules in decreasing order of
a) average KE
b) rms speeds
a) Pressure and temperature are the same, therefore,
KEc > KEb > KEa
b) When P and T are constant, (Vrms)c > (Vrms)b > (Vrms)a
We have 0.5 g of hydrogen gas in a cubic chamber of size 3 cm kept at NTP. The gas in the chamber is compressed, keeping the temperature constant till a final pressure of 100 atm. Is one justified in assuming the ideal gas law, in the final state?
Volume of 1 molecule = 4/3 πr3 = 4.20 × 10-30 m3
No.of moles in 0.5 g H2 gas = 0.25 mole
Volume of H2 molecule in 0.25 mole = 1.04×6.023× 10+23-30
When air is pumped into a cycle tyre, the volume and pressure of the air in the tyre both are increased. What about Boyle’s law in this case?
PV = P(m/ρ) = constant
P/ ρ = constant
Volume = m/ ρ where m is constant
Therefore, when the air is pumped into the tyre of the cycle, the mass of air increases as the no.of molecules increases. Therefore, Boyle’s law is only applicable when the mass of the gas remains fixed.
A balloon has 5.0 g mole of helium at 7oC. Calculate
a) the number of atoms of helium in the balloon
b) the total internal energy of the system
Average KE per molecule = 3/2kT
No.of moles of helium, n = 5 g mole
T = 280 K
a) No.of atoms of helium in the balloon = 30.015 × 1023
b) KE = 3/2 kbT
Total internal energy = 1.7 × 104J
Calculate the number of degrees of freedom of molecules of hydrogen in 1 cc of hydrogen gas at NTP.
The volume occupied by the molecules of gas = 22400 cc
No.of molecules in 1 cc of hydrogen = 2.688 × 1019
Hydrogen has a total of 5 degrees of free as it is a diatomic gas
Therefore, the total degrees of freedom = 1.344 × 1020
An insulated container containing monoatomic gas of molar mass m is moving with a velocity vo. If the container is suddenly stopped, find the change in temperature.
The final KE of the gas = 0
Change in KE, ∆K= 1/2 (nm)v2
∆T is the change in the temperature
∆U = nCv∆T
∆K = ∆U
∆T = mv02/3R
Long AnswersExplain why
a) there is no atmosphere on moon
b) there is a fall in temperature with altitude
a) There is no atmosphere on the moon because the gravitational force is small, and the Vrms is greater on the moon such that the escape velocity of the molecule is greater than that of the air. Also, the distance between the moon and the sun is the same as the distance between the moon and the earth. Therefore, the RMS speed of the molecule increases such that it can be more than the escape velocity. Therefore, there is no atmosphere on the moon.
b) There is a fall in temperature with altitude because at the higher altitude, the gas molecules expand, and there is a decrease in temperature as the expansion of gas results in cooling of the surrounding. This results in an increase in the kinetic energy of the molecules. When the kinetic energy increases, temperature decreases.
Consider an ideal gas with the following distribution of speeds
| Speed (m/s) | % of molecules |
| 200 | 10 |
| 400 | 20 |
| 600 | 40 |
| 800 | 20 |
| 1000 | 10 |
a) calculate Vrms and hence T (m = 3.0 × 10-26 kg)
b) if all the molecules with speed 1000 m/s escape from the system, calculate new Vrms and hence T
a) T = 296 K
b) T = 248.04 K
Ten small planners are flying at a speed of 150 km/h in total darkness in an air space that is 20 × 20 × 1.5 km3 in volume. You are in one of the planes, flying at random within this space with no way of knowing where the other planes are. On average about how long a time will elapse between near collision with your plane? Assume for this rough computation that a safety region around the plane can be approximated by a sphere of radius 10 m.
Time = distance/speed
No.of particles per unit volume v = N/volume
n = 0.0167 km-3
d = 10 × 10-3 km
v = 150 km/hr
Therefore, time = 225 hrs
A box of 1.00 m3 is filled with nitrogen at 1.50 atm at 300 K. The box has a hole of an area 0.010 mm2. How much time is required for the pressure to reduce by 0.10 atm, if the pressure outside is 1 atm?
Volume of the box = 1 m3 = V1
Initial pressure P1 = 1.5 atm
Final pressure P2 = 1.4 atm
Air pressure Pa = 1 atm
Initial temperature T1 = 300 K
Final temperature T2 = 300 K
Area of the hole = 10-8 m2
Pressure difference between tyre and atmosphere = 1.5 – 1 atm
Mass of nitrogen = 46.5 × 10-27 kg
Using the above information, we can calculate the time required as τ = 1.34 × 105 sec