Q. No.To find out the degree of freedom, the correct expression is:
1. \(f=\frac{2}{\gamma -1}\)
2. \(f=\frac{\gamma+1}{2}\)
3. \(f=\frac{2}{\gamma +1}\)
4. \(f=\frac{1}{\gamma +1}\)
1. \(f=\frac{2}{\gamma -1}\)
2. \(f=\frac{\gamma+1}{2}\)
3. \(f=\frac{2}{\gamma +1}\)
4. \(f=\frac{1}{\gamma +1}\)
The equation of state for 5g of oxygen at a pressure P and temperature T, when occupying a volume V, will be: (where R is the gas constant)
1. PV = 5 RT
2. PV = (5/2) RT
3. PV = (5/16) RT
4. PV = (5/32) RT
Uranium has two isotopes of masses \(235 \) and \(238\) units. If both are present in Uranium hexafluoride gas, which would have the larger average speed?
1. \(^{235} \mathrm{U} \mathrm{F}_{6}\)
2. \({}^{238} \mathrm{U} \mathrm{F}_{6}\)
3. Both will have the same average speed.
4. Data insufficient
| 1. | \(379\) J | 2. | \(357\) J |
| 3. | \(457\) J | 4. | \(374\) J |
Match Column-I and Column-II and choose the correct match from the given choices.
| Column-I | Column-II | ||
| (A) | Root mean square speed of gas molecules | (P) | \(\dfrac13nm\bar v^2\) |
| (B) | The pressure exerted by an ideal gas | (Q) | \( \sqrt{\dfrac{3 R T}{M}} \) |
| (C) | The average kinetic energy of a molecule | (R) | \( \dfrac{5}{2} R T \) |
| (D) | The total internal energy of a mole of a diatomic gas | (S) | \(\dfrac32k_BT\) |
| (A) | (B) | (C) | (D) | |
| 1. | (Q) | (P) | (S) | (R) |
| 2. | (R) | (Q) | (P) | (S) |
| 3. | (R) | (P) | (S) | (Q) |
| 4. | (Q) | (R) | (S) | (P) |
An increase in the temperature of a gas-filled in a container would lead to:
| 1. | decrease in the intermolecular distance. |
| 2. | increase in its mass. |
| 3. | increase in its kinetic energy. |
| 4. | decrease in its pressure. |
If \(C_P\) and \(C_V\) denote the specific heats (per unit mass) of an ideal gas of molecular weight \(M\) (where \(R\) is the molar gas constant), the correct relation is:
1. \(C_P-C_V=R\)
2. \(C_P-C_V=\frac{R}{M}\)
3. \(C_P-C_V=MR\)
4. \(C_P-C_V=\frac{R}{M^2}\)
The mean free path \(l\) for a gas molecule depends upon the diameter, \(d\) of the molecule as:
| 1. | \(l\propto \dfrac{1}{d^2}\) | 2. | \(l\propto d\) |
| 3. | \(l\propto d^2 \) | 4. | \(l\propto \dfrac{1}{d}\) |
Diatomic molecules like hydrogen have energies due to both translational as well as rotational motion. The equation in kinetic theory \(PV = \dfrac{2}{3}E,\) \(E\) is:
| 1. | the total energy per unit volume. |
| 2. | only the translational part of energy because rotational energy is very small compared to translational energy. |
| 3. | only the translational part of the energy because during collisions with the wall, pressure relates to change in linear momentum. |
| 4. | the translational part of the energy because rotational energies of molecules can be of either sign and its average over all the molecules is zero. |
\(1\) mole of an ideal gas is contained in a cubical volume V, ABCDEFGH at \(300\) K (figure). 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:
| 1. | the pressure on EFGH would be zero. |
| 2. | the pressure on all the faces will be equal. |
| 3. | the pressure on EFGH would be double the pressure on ABCD. |
| 4. | the pressure on EFGH would be half that on ABCD. |
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