4 Marks Question

Question 14 Marks

50cc of oxygen is collected in an inverted gas jar over water. The atmospheric pressure is 99.4kPa and the room temperature is 27°C. The water level in the jar is same as the level outside. The saturation vapour pressure at 27°C is 3.4kPa. Calculate the number of moles of oxygen collected in the jar.

Answer

Pressure inside the tube = Atmospheric Pressure = 99.4KPa

Pressure exerted by O₂ vapour = Atmospheric pressure - V.P.

= 99.4KPa - 3.4KPa = 96KPa

No of moles of O₂ = n

$96\times10^3\times50\times10^{-6}=\text{n}\times8.3\times300$

$\Rightarrow\text{n}=\frac{96\times50\times10^{-3}}{8.3\times300}=1.9277\times10^{-3}\approx1.93\times10^{-3}$

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Question 24 Marks

On a winter day, the outside temperature is 0°C and relative humidity 40%. The air from outside comes into a room and is heated to 20°C. What is the relative humidity in the room? The saturation vapour pressure at 0°C is 4.6mm of mercury and at 20°C it is 18mm of mercury.

Answer

Relative humidity = 40%

SVP = 4.6mm of Hg

$0.4=\frac{\text{VP}}{4.6}\Rightarrow\text{VP}=0.4\times4.6=1.84$

$\frac{\text{P}_1\text{V}}{\text{T}_1}=\frac{\text{P}_2\text{V}}{\text{T}_2}\Rightarrow\frac{1.84}{273}=\frac{\text{P}_2}{293}$

$\Rightarrow\text{P}_2=\frac{1.84}{273}\times293$

Relative humidity at 20°C

$=\frac{\text{VP}}{\text{SVP}}=\frac{1.84\times293}{273\times10}=0.109=10.9\%$

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Question 34 Marks

Read the passage given below and answer the following questions from (i) to (v).

Monatomic Gases: The molecule of a monatomic gas has only three translational degrees of freedom. Thus, the average energy of a molecule at temperature T is $\frac{3}{2}\text{K}_\text{b}\text{T}$. The total internal energy of a mole of such a gas is $\text{U}=(\frac{3}{2})\text{RT}$.

The molar specific heat at constant volume cv is given by

$\text{C}_{\text{v}}=\frac{\text{Du}}{\text{Dt}}=(\frac{3}{2})\text{R}$

For an ideal gas,

C_p - C_v = R

Where Cp is the molar specific heat at constant pressure. Thus, $\text{C}_\text{P} =(\frac{5}{2})\text{R}$

The ratio of specific heats IS $\gamma=\frac{\text{cp}}{\text{cv}}=\frac{5}{3}$

Diatomic Gases: a diatomic molecule treated as a rigid rotator, like a dumbbell, has 5 degrees of freedom: 3 translational and 2 rotational. Using the law of equipartition of energy, the total internal energy of a mole of such a gas is $\text{U}=\frac{5}{2}\text{RT}$

The molar specific heat at constant volume cv is given by

$\text{Cv}=\frac{\text{DU}}{\text{DT}}=(\frac{5}{2})\text{R}$

For an ideal gas,

C_p – C_v= R

Where Cp is the molar specific heat at constant pressure. Thus, $\text{C}_\text{P} =(\frac{7}{2})\text{R}$

The ratio of specific heats IS $γ( \text{for rigid diatomic)}=\frac{\text{C}_\text{P}}{\text{C}_\text{v}} =(\frac{7}{5})\text{R}$

For non rigid diatomic molecules they have additional mode of vibrations therefore

$\gamma=\frac{\text{C}_\text{p}}{\text{C}_\text{v}}=\frac{9}{7}$

Polyatomic Gases: In general a polyatomic molecule has 3 translational, 3 rotational degrees of freedom and a certain number (f) of vibrational modes. According to the law of equipartition of energy, it is easily seen that one mole of such a gas has

C_v= (3 + f) R and C_p= (4 + f) R and $\gamma=\frac{(4 + \text{f})}{(3+\text{f})}$

For monatomic molecules ratio of specific heats is $\gamma$

$\frac{5}{3}$
$\frac{7}{5}$
$\frac{9}{5}$
None of these

For diatomic rigid molecules ratio of specific heats is γ

$\frac{5}{3}$
$\frac{7}{5}$
$\frac{9}{7}$
None of these

For diatomic non rigid molecules ratio of specific heats is γ

$\frac{5}{3}$
$\frac{7}{5}$
$\frac{9}{7}$
None of these

Give cp and cv values and ratio of specific heat for monatomic gas molecules.
Give cp and cv values and ratio of specific heat for polyatomic gas molecules

Answer

(a)$\frac{5}{3}$
(b)$\frac{7}{5}$
(c)$\frac{9}{7}$
Monatomic Gases: The molecule of a monatomic gas has only three translational degrees of freedom. Thus, the average energy of a molecule at temperature T is $\frac{3}{2}\text{K}_\text{b}\text{T}$. The total internal energy of a mole of such a gas is $\text{U}=(\frac{3}{2})\text{RT}$. The molar specific heat at constant volume c_v is given by $\text{C}_{\text{v}}=\frac{\text{Du}}{\text{Dt}}=(\frac{3}{2})\text{R}$ For an ideal gas, C_p – C_v = R Where C_p is the molar specific heat at constant pressure. Thus, $\text{C}_\text{P} =(\frac{5}{2})\text{R}$ The ratio of specific heats IS $\gamma=\frac{\text{cp}}{\text{cv}}=\frac{5}{3}$
Polyatomic Gases: In general a polyatomic molecule has 3 translational, 3 rotational degrees of freedom and a certain number (f) of vibrational modes. According to the law of equipartition of energy, it is easily seen that one mole of such a gas has C_v= (3+f) R and C_p= (4+f) R and $\gamma=\frac{(4 + \text{f})}{(3+\text{f})}$

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Question 44 Marks

The human body has an average temperature of 98°F. Assume that the vapour pressure of the blood in the veins behaves like that of pure water. Find the minimum atmospheric pressure which is necessary to prevent the blood from boiling. Use figure. of the text for the vapour pressures.

Answer

The temp. of body is 98°F = 37°C
At 37°C from the graph SVP = Just less than 50mm
B.P. is the temp. when atmospheric pressure equals the atmospheric pressure.
Thus min. pressure to prevent boiling is 50mm of Hg.

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Question 54 Marks

Read the passage given below and answer the following questions from (i) to (v).
Root mean square velocity (RMS value)is the square root of the mean of squares of the velocity of individual gas molecules and the Average velocity is the arithmetic mean of the velocities of different molecules of a gas at a given temperature.

Moon has no atmosphere because:-
1. It is far away from the surface of the earth
2. Its surface temperature is 10°C
3. The r.m.s. velocity of all the gas molecules is more than the escape velocity of the moon’s surface
4. The escape velocity of the moon’s surface is more than the r.m.s velocity of all molecules
For an ideal gas, CPCV is
1. >1
2. <1
3. ≤1
4. none of these
The root mean square velocity of hydrogen is 5 times than that of nitrogen. If T is the temperature of the gas then:
1. T(H₂) = T(N₂)
2. T(H₂) < T(N₂)
3. T(H₂) > T(N₂)
4. none of these
Suppose the temperature of the gas is tripled and N₂ molecules dissociate into an atom. Then what will be the rms speed of atom:
1. v06
2. v0
3. v03
4. none of these
The velocities of the molecules are v, 2v, 3v, 4v & 5v. The rms speed will be:
1. 11 v
2. v(11)¹²
3. v
4. v(12)¹¹

Answer

(c) The r.m.s. velocity of all the gas molecules is more than the escape velocity of the moon’s surface
(b) <1
(b) T(H₂) < T(N₂)
(a) v06
(b) v(11)¹²

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Question 64 Marks

Read the passage given below and answer the following questions from (i) to (v).

Boyle’s law is a gas law which states that the pressure exerted by a gas (of a given mass, kept at a constant temperature) is inversely proportional to the volume occupied by it. In other words, the pressure and volume of a gas are inversely proportional to each other as long as the temperature and the quantity of gas are kept constant. For a gas, the relationship between volume and pressure (at constant mass and temperature) can be expressed mathematically as follows.

$\text{P}\infty(\frac{1}{\text{V}})$ Where P is the pressure exerted by the gas and V is the volume occupied by it. This proportionality can be converted into an equation by adding a constant, k.

Charles law states that the volume of an ideal gas is directly proportional to the absolute temperature at constant pressure. The law also states that the Kelvin temperature and the volume will be in direct proportion when the pressure exerted on a sample of a dry gas is held constant. Charles law and Boyle’s law applied to low density gas only. The total pressure of a mixture of ideal gases is the sum of partial pressures. This is Dalton’s law of partial pressures.

Boyle’s law is obeyed by high as well as low density gases. True or False?

True
False

Charles law is states that volume of an ideal gas is directly proportional to temperature at constant

Temperature
Pressure
Volume
None of these

State Daltons law of partial pressures
State Boyle’s law
State Charles law

Answer

(a) True
(b) Pressure
The total pressure of a mixture of ideal gases is the sum of partial pressures exerted by all the molecules of gas. This is Dalton’s law of partial pressures.
Boyle’s law is a gas law which states that at constant temperature the pressure exerted by a gas is inversely proportional to the volume occupied by it. In other words, the pressure and volume of a gas are inversely proportional to each other as long as the temperature and the quantity of gas are kept constant. For a gas, the $\text{P}\infty(\frac{1}{\text{V}})$ Where P is the pressure exerted by the gas and V is the volume occupied by it. This proportionality can be converted into an equation by adding a constant k.
Charles law states that the volume of an ideal gas is directly proportional to the absolute temperature at constant pressure.

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Question 74 Marks

Read the passage given below and answer the following questions from (i) to (v).

Pressure of an Ideal Gas: according to kinetic theory of gases pressure is given by

$\text{P}=\frac{1}{3}\text{ nmv}^2$

Where, n is number of molecules per unit volume, m is mass and v² is mean squared speed. Though we choose the container to be a cube, the shape of the vessel really is immaterial.

The average kinetic energy of a molecule is proportional to the absolute temperature of the gas; it is independent of pressure, volume or the nature of the ideal gas. This is a fundamental result relating temperature, a macroscopic measurable parameter of a gas (a thermodynamic variable as it is called) to a molecular quantity, namely the average kinetic energy of a molecule. The two domains are connected by the Boltzmann constant and given by E = k_bT.

Where kb is Boltzmann constant having value of 1.38 × 10^-23 joule per Kelvin.

We have seen that in thermal equilibrium at absolute temperature T, for each translational mode of motion, the average energy is $\frac{1}{2}\text{K}_\text{b}\text{t}$. The most elegant principle of classical statistical mechanics (first proved by Maxwell) states that this is so for each mode of energy: translational, rotational and vibrational. That is, in equilibrium, the total energy is equally distributed in all possible energy modes, with each mode having an average energy equal to $\frac{1}{2}\text{K}_\text{b}\text{t}$. This is known as the law of equipartition of energy. Accordingly, each translational and rotational degree of freedom of a molecule contributes $\frac{1}{2}\text{K}_\text{b}\text{t}$ to the energy, while each vibrational frequency contributes $2\times\frac{1}{2}\text{Kb T}=\text{K}_\text{b}\text{T}$ since a vibrational mode has both kinetic and potential energy modes.

Boltzmann constant has value of:

1.38 × 10 - 23 joule per Kelvin.
1.38 × 10 - 28 joule per Kelvin.
1.38 × 10 - 30 joule per Kelvin.
None of these.

SI unit of Boltzmann constant is given by:

Joules per meter
Joules per Kelvin
Joules per Newton
None of these

According to kinetic theory give formula for pressure of idea gas.
According to kinetic theory what is average kinetic energy of molecules of ideal gas?
What is law of equipartition of energy?

Answer

(a) 1.38 × 10 - 23 joule per Kelvin.
(b) Joules per Kelvin
According to kinetic theory of gases pressure is given by $\text{P}=\frac{1}{3}\text{ nmv}^2$ Where, n is number of molecules per unit volume, m is mass and v² is mean squared speed. Though we choose the container to be a cube, the shape of the vessel really is immaterial.
The average kinetic energy of a molecule is proportional to the absolute temperature of the gas; it is independent of pressure, volume or the nature of the ideal gas and given by $\text{E}=\frac{3}{2}\text{K}_\text{b}\text{t}$. Where kb is Boltzmann constant having value of 1.38 × 10 - 23 joule per Kelvin.
We know that for each translational mode of motion, the average energy is $\frac{1}{2}\text{K}_\text{b}\text{t}$. classical statistical mechanics states that in equilibrium, the total energy is equally distributed in all possible energy modes, with each mode having an average energy equal to $\frac{1}{2}\text{K}_\text{b}\text{t}$. This is known as the law of equipartition of energy. Accordingly, each translational and rotational degree of freedom of a molecule contributes $\frac{1}{2}\text{K}_\text{b}\text{t}$ to the energy, while each vibrational frequency contributes $2\times\frac{1}{2}\text{Kb T}=\text{K}_\text{b}\text{T}$ since a vibrational mode has both kinetic and potential energy modes.

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Physics 4 Marks Question

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