The root mean square speed of molecules of a given mass of a gas at \(27^\circ \text{C}\) and \(1~\text{atm}\) is \(200~\text{m/s}\) The root mean square speed of molecules of the gas at \(127^\circ \text{C}\) and \(2~\text{atm}\) will be:

1. \(\dfrac{200}{\sqrt{3}}~\text{m/s}\) 2. \(\dfrac{200}{\sqrt{5}}~\text{m/s}\)
3. \(\dfrac{400}{\sqrt{3}}~\text{m/s}\) 4. \(\dfrac{100}{\sqrt{5}}~\text{m/s}\)

Subtopic:  Types of Velocities |
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Molecules of an ideal gas are known to have three translational degrees of freedom and two rotational degrees of freedom. The gas is maintained at a temperature of \(T.\) The total internal energy, \(U\) of a mole of this gas, and the value of \(\gamma~\left(=\dfrac{C_P}{C_V}\right )\) are, respectively:
1. \( U=5 R T \text { and } \gamma=\dfrac{7}{5} \)

2. \( U=\dfrac{5}{2} R T \text { and } \gamma=\dfrac{6}{5} \)

3. \(U=5 R T \text { and } \gamma=\dfrac{6}{5} \)

4. \( U=\dfrac{5}{2} R T \text { and } \gamma=\dfrac{7}{5}\)

Subtopic:  Law of Equipartition of Energy |
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Initially, a gas of diatomic molecules is contained in a cylinder of volume \(V_1\) at a pressure \(P_1\) and temperature \(250~\text{K}.\) Assume that \(25\%\) of the molecules get dissociated, causing a change in the number of moles. The pressure of the resulting gas at temperature \(2000~\text{K},\) when contained in a volume \(2V_1\) is given by \(P_2.\) The ratio \(P_2/P_1\) is:
1. \(2\)
2. \(3\)
3. \(5\)
4. \(9\)

Subtopic:  Ideal Gas Equation |
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Level 3: 35%-60%
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The temperature of an open room of volume \(30~\text{m}^3\) increases from \(17^\circ \text{C}\) to \(27^\circ \text{C}\) due to the sunshine. The atmospheric pressure in the room remains \(1\times 10^{5}~\text{Pa}\). In \(n_i\) and \(n_f\) are the number of molecules in the room before and after heating, the \(n_f\text-n_i \) will be:
1. \( -1.61 \times 10^{23} \)
2. \( 1.38 \times 10^{23} \)
3. \( 2.5 \times 10^{25} \)
4. \( -2.5 \times 10^{25}\)

Subtopic:  Ideal Gas Equation |
Level 3: 35%-60%
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The internal energy (\(U\)), pressure (\(P\)), and volume (\(V\)) of an ideal gas are related as \(U=3PV+4\). The gas is:

1. Diatomic only
2. Polyatomic only
3. Either monoatomic or diatomic
4. Monoatomic only

Subtopic:  Law of Equipartition of Energy |
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On the basis of kinetic theory of gases, the gas exerts pressure because its molecules:

1. continuously lose their energy till it reaches wall.
2. are attracted by the walls of container.
3. continuously stick to the walls of container.
4. suffer change in momentum when impinge on the walls of container.

Subtopic:  Ideal Gas Equation |
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If the ratio of the number density per cm3 of the two gases is \(5:3\) and the ratio of the diameters of the molecules of the two gases is \(4:5,\) then, the ratio of the mean free path of molecules of two gases is:

1. \(\dfrac{16}{15}\) 2. \(\dfrac{15}{16}\)
3. \(\dfrac{3}{4}\) 4. \(\dfrac{4}{3}\)
Subtopic:  Mean Free Path |
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Given below are two statements :

Statement I: In a diatomic molecule, the rotational energy at a given temperature obeys Maxwell's distribution.
Statement II: In a diatomic molecule, the rotational energy at a given temperature equals the translational kinetic energy for each molecule.

In the light of the above statements, choose the correct answer from the options given below :
 

1. Statement I is false but Statement II is true.
2. Both Statement I and Statement II are false.
3. Both Statement I and Statement II are true.
4. Statement I is true but Statement II is false.

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Match the \(C_p/C_V\)  ratio for ideal gases with different types of molecules:

Column I Column II
(A) Monatomic (I) \(7/5\)
(B) Diatomic rigid molecules (II) \(9/7\)
(C) Diatomic non-rigid molecules (III) \(4/3\)
(D) Triatomic rigid molecules (IV) \(5/3\)
 
1. (A)-(III), (B)-(IV), (C)-(II), (D)-(I)
2. (A)-(II), (B)-(III), (C)-( I), (D)-(IV)
3. (A)-(IV), (B)-(II), (C)-(I), (D)-(III)
4. (A)-(IV), (B)-(I), (C)-(II), (D)-(III)
 

Subtopic:  Law of Equipartition of Energy |
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Nitrogen gas is at a certain temperature \(300^\circ \text{C}.\) At what temperature (in Kelvin) will the root mean square (rms) speed of a hydrogen molecule be equal to the rms speed of a nitrogen molecule?
(given: molar mass of nitrogen molecule is \(28~\text g/ \text{mol}\) and molar mass of hydrogen molecule is \(2~\text g/ \text{mol}\))
1. \(21\) K
2. \(41\) K
3. \(52\) K
4. \(76\) K

Subtopic:  Types of Velocities |
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