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Practice each question, then open the answer or worked solution.
Ina 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
The correct answer is
a) obeys Maxwell’s distribution
d) is 2/3rd the translational kinetic energy for each molecule
Which of the following diagrams depicts ideal gas behaviour?
The correct answer is
a)
c)
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 the redistribution of energy amongst the molecules
The correct answer is a) because of collisions with moving parts of the wall only
Very Short AnswersCalculate the number of atoms in 39.4 g gold. The molar mass of gold is 197 g/mole.
Molar mass = 6.023 × 1023 atoms
Mass of gold, m = 39.4 g
Molar mass of gold, M = 197 g/mol
196 g of gold contains 6.023 × 1023 atoms
39.4 g of gold contains 1.20 × 1023 atoms.
The volume of a given mass of a gas at 27oC, 1 atm is 100 cc. What will be its volume at 327oC?
T1 = 27oC
T1 = 300 K
V1 = 100 cm3
We know that V is proportional to T
V/T = constant
V1/T1 = V2/T2
V2 = V1(T2/T1)
V2 = 200 cc
The molecules of a given mass of 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?
Vrms = 100 m/s
T1 = 300 K
T2 = 400 K
Vrms = √3RT/M
V2rms =?
Vrms = 115.4 m/s
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?
V1 = 9 × 106 m/s
V2 = 1 × 106 m/s
Vrms = 6.4 × 106 m/s
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.)
Degree of freedom of oxygen = 5
Total internal energy of 2-mole oxygen = 5RT
Total internal energy of 4 mole of Ne = 6RT
The total internal energy of the system = 11 RT
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.
The ratio of the mean free paths of the molecules is 4:1.
Short AnswersThe 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 |
For an ideal gas, PV = μRT
P1V1 = μ1R1T1
P2V2 = μ2R2T2
P1 = 1 atm
P2 = 2 atm
V1 = 2L
V2 = 3L
T1 = T = T2
μ1 = 4
μ2 = 5
Substituting all the values we get, P = 1.6 atm