, where m is the mass of the gas molecule. Answer: Let ρm represent the number of molecules per unit volume Then the expression for the change in momentum by a molecule on front side is = 2m (v +...
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?
Answer: 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 =...
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 the 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.
Answer: We know that, Time = distance/speed Number of particles per unit volume v = N/volume n = 0.0167 km-3 d = 10 × 10-3 km v = 150 km/hr Therefore, we get: time = 225 hrs
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...
Explain why
a) there is no atmosphere on moon b) there is a fall in temperature with altitude Answer: a) The moon has no atmosphere since the gravitational pull is minimal and the Vrms is bigger on the moon,...
Answer: According to the question, the final KE of the gas is 0 The change in KE is as follows: ∆K= 1/2 (nm)v2 ∆T is the change in the temperature ∆U = nCv∆T ∆K = ∆U Making use of the expression, we...
Calculate the number of degrees of freedom of molecules of hydrogen in 1 cc of hydrogen gas at NTP.
Answer: According to the question, the volume occupied by the molecules of a gas is 22400 cc and the number of molecules in 1 cc of hydrogen are 2.688 × 1019 We know that the hydrogen has a total...
A balloon has 5.0 g mole of helium at 7 degrees C. Calculate
a) the number of atoms of helium in the balloon b) the total internal energy of the system Answer: We know that the average KE per molecule is 3/2kT No.of moles of helium are n = 5 g mole We have T...
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?
Answer: We have, PV = P(m/ρ) = constant P/ ρ = constant Volume = m/ ρ where m is constant When air is pushed into the cycle's tyre, the mass of the air increases as the number of molecules...
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?
Answer: We know that volume of 1 molecule = 4/3 πr3 = 4.20 × 10-30 m3 And the number 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 = 6.264 ×...
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 Answer: a) From the above result, the pressure and temperature are the same, therefore, KEc > KEb > KEa b) When P and T are constant, (Vrms)c > (Vrms)b >...
The container shown in the figure has two chambers, separated by a partition, of volumes V1 = 2.0 litre and V2 = 3.0 litre. The chambers contain μ1 = 4.0 and μ2 = 5.0 moles of a gas at pressure p1 = 1.00 atm and p2 = 2.00 atm. Calculate the pressure after the partition is removed and the mixture attains equilibrium.
V1 V2 μ1 μ2 p1 p2 Answer: We know that for an ideal gas, PV = μRT The two equations that we can write are: P1V1 = μ1R1T1 P2V2 = μ2R2T2 P1 = 1 atm, P2 = 2 atm and V1 = 2L, V2 = 3L Also, T1 = T = T2...
Calculate the ratio of the mean free paths of the molecules of two gases having molecular diameters 1 A and 2 A. The gases may be considered under identical conditions of temperature, pressure, and volume.
Answer: We know that that we can write: I α (1/d2) d1 = 1Å d2 = 2Å Therefore, l1 : l2 = 4 : 1 The ratio of the mean free paths of the molecules is 4:1
A gas mixture consists of 2.0 moles of oxygen and 4.0 moles of neon a temperature T. Neglecting all vibrational modes, calculate the total internal energy of the system. (Oxygen has two rotational modes.)
Answer: We know that oxygen has 5 degrees of freedom. Therefore, energy per mole = (5/2)RT Therefore, for 2 moles of oxygen, energy = 5RT Neon has 3 degrees of freedom. Therefore, energy per mole...
Two molecules of a gas have speeds of 9 × 106 m/s and 1 × 106 m/s respectively. What is the root mean square speed of these molecules?
Answer: Expression for root mean square velocity is as follows: $ {{v}_{rms}}=\sqrt{\frac{v_{1}^{2}+v_{2}^{2}+v_{3}^{2}+......+v_{n}^{2}}{n}} $ For two meolecules, the formula reduces to: $...
The molecules of a given mass of a gas have root mean square speeds of 100 m/s at 27oC and 1.00 atmospheric pressure. What will be the root mean square speeds of the molecules of the gas at 127oC and 2.0 atmospheric pressure?
Answer: According to the question, Vrms = 100 m/s T1 = 300 K and T2 = 400 K Expression for root mean square velocity is Vrms = √3RT/M Upon substituting the given values, we get Vrms = 115.4...
The volume of a given mass of a gas at 27 degrees C, 1 atm is 100 cc. What will be its volume at 327oC?
Answer: According to the question, T1 = 27oC = 300 K and the volume V1 = 100 cm3 It is known that V is proportional to T So, we can write V/T = constant Or, V1/T1 = V2/T2 Upon re-arranging, we have...
Calculate the number of atoms in 39.4 g gold. Molar mass of gold is 197 g/mole.
Answer: According to the question, the molar mass = mass of Avagadro's number of atoms = 6.023 × 1023 atoms Mass of gold is m = 39.4 g and the molar mass of gold is M = 197 g/mol We know that 197 g...
When an ideal gas is compressed adiabatically, its temperature rises: the molecules on the average have more kinetic energy than before. The kinetic energy increases,
a) because of collisions with moving parts of the wall only b) because of collisions with the entire wall c) because the molecules gets accelerated in their motion inside the volume d) because of...
Which of the following diagrams depicts ideal gas behaviour?
Answer: The correct answer is a) c)
In a diatomic molecules, the rotational energy at a given temperature
a) obeys Maxwell’s distribution b) have the same value for all molecules c) equals the translational kinetic energy for each molecule d) is 2/3rd the translational kinetic energy for each molecule...
Diatomic molecules like hydrogen have energies due to both translational as well as rotational motion. From the equation in kinetic theory pV = 2/3 E, E is
a) the total energy per unit volume d) only the translational part of energy because rotational energy is very small compared to the translational energy c) only the translational part of the energy...
ABCDEFGH is a hollow cube made of an insulator. Face ABCD has a positive charge on it. Inside the cube, we have ionized hydrogen. The usual kinetic theory expression for pressure
a) will be valid b) will not be valid since the ions would experience forces other than due to collision with the walls c) will not be valid since collisions with walls would not be elastic d) will...
An inflated rubber balloon contains one mole of an ideal gas, has a pressure p, volume V, and temperature T. If the temperature rises to 1.1T and the volume is increased to 1.05V, the final pressure will be
a) 1.1 p b) p c) less than p d) between p and 1.1 Answer: The correct option is d) between p and 1.1 Explanation:
A vessel of volume V contains a mixture of 1 mole of hydrogen and 1 mole of oxygen. Let f1(v)dv, denote the fraction of molecules with speed between v and (v + dv) with f2(v)dv similarly for oxygen. Then
a) f1(v) + f2(v) = f(v) obeys the Maxwell’s distribution law b) f1(v), f2(v) will obey the Maxwell’s distribution law separately c) neither f1(v) nor f2(v) will obey the Maxwell’s distribution law...
1 mole of H2 gas is contained in a box of volume V = 1.00 m3 at T = 300 K. The gas is heated to a temperature of T = 3000 K and the gas gets converted to a gas of hydrogen atoms. The final pressure would be
a) same as the pressure initially b) 2 times the pressure initially c) 10 times the pressure initially d) 20 times the pressure initially Answer: The correct option is d) 20 times the pressure...
Volume versus temperature graphs for a given mass of an ideal gas are shown in the figure at two different values of constant pressure. What can be inferred about the relation between P1 and P2?
a) P1 > P2 b) P1 = P2 c) P1 < P2 d) data is insufficient Answer: The correct option is a) P1 > P2 Explanation: When the pressure of an ideal gas is constant, Chale's law is obeyed, i.e. V ∝...
A cylinder containing an ideal gas is in a vertical position and has a piston of mass M that is able to move up or down without friction. If the temperature is increased,
a) both p and V of the gas will change b) only p will increase according to Charle’s law c) V will change but not p d) p will change but not V Answer: The correct option is c) V will change but not...
Boyle’s law is applicable for an
a) adiabatic process b) isothermal process c) isobaric process d) isochoric process Answer: The correct option is b) isothermal process Explanation: In an isothermal process, when the temperature...
1 mole of an ideal gas is contained in a cubical volume V, ABCDEFGH at 300 K. One face of the cube (EFGH) is made up of a material which totally absorbs any gas molecule incident on it. At any given time,
a) the pressure on EFGH would be zero b) the pressure on all the faces will the equal c) the pressure of EFGH would be double the pressure on ABCD d) the pressure on EFGH would be half that on ABCD...
A cubic vessel (with faces horizontal + vertical) contains an ideal gas at NTP. The vessel is being carried by a rocket which is moving at a speed of 500 m/s in the vertical direction. The pressure of the gas inside the vessel as observed by us on the ground
a) remains the same because 500 m/s is very much smaller than vrms of the gas b) remains the same because the motion of the vessel as a whole does not affect the relative motion of the gas molecules...