The work done by 2 moles of polyatomic gas \( \gamma = \frac{4}{3} \) initially at room temperature to increase its volume eight time during adiabatic process will be (Take \(R = 2\text{ cal mol}^{-1}\text{ K}^{-1}\) and room temperature 27°C)
1. 900 cal
2. 600 cal
3. 1800 cal
4. 1200 cal
View Answer
In adiabatic process, \(T_1 V_1^{\gamma - 1} = T_2 V_2^{\gamma - 1}\). Here, \(\gamma - 1 = 1/3\). Since \(V_2/V_1 = 8\), we get \(T_2 = T_1 (1/8)^{1/3} = T_1/2 = 300/2 = 150\text{ K}\). The work done is \(W = \frac{nR(T_1 - T_2)}{\gamma - 1} = \frac{2 \times 2 \times (300 - 150)}{1/3} = 1800\text{ cal}\).
A gas mixture consists of 5 moles of oxygen and 3 moles of argon at temperature T. Assuming the gases to be ideal and oxygen bond to be rigid, the total internal energy of the mixture is (where R denotes the universal gas constant)
1. 17RT
2. 15RT
3. 20RT
4. 11RT
View Answer
Internal energy is \( U = n \frac{f}{2} RT \). For diatomic \( \text{O}_2 \) (rigid, \( f_1=5 \)), \( U_1 = 5 \times \frac{5}{2} RT = 12.5 RT \). For monoatomic \( \text{Ar} \) (\( f_2=3 \)), \( U_2 = 3 \times \frac{3}{2} RT = 4.5 RT \). Total \( U = 12.5RT + 4.5RT = 17RT \).
The ratio of \(v_{\text{rms}} : v_{\text{mp}} : v_{\text{avg}}\) is (symbols have their usual meaning)
1. \(\sqrt{3} : \sqrt{2} : \sqrt{\frac{\pi}{8}}\)
2. \(\sqrt{3} : \sqrt{2} : \sqrt{\frac{8}{3}}\)
3. \(\sqrt{3} : \sqrt{\frac{8}{\pi}} : \sqrt{2}\)
4. \(\sqrt{3} : \sqrt{2} : \sqrt{\frac{8}{\pi}}\)
View Answer
Using standard molecular velocity expressions: \(v_{\text{rms}} = \sqrt{\frac{3RT}{M}}\), \(v_{\text{mp}} = \sqrt{\frac{2RT}{M}}\), and \(v_{\text{avg}} = \sqrt{\frac{8RT}{\pi M}}\). This gives the ratio \(\sqrt{3} : \sqrt{2} : \sqrt{\frac{8}{\pi}}\).
Assertion (A): Density of humid air is less then density of dry air at the same temperature and pressure.
Reason (R): Mass of humid air is more than mass of dry air.
1. Both (A) & (R) are true and the (R) is the correct explanation of the (A)
2. Both (A) & (R) are true but the (R) is not the correct explanation of the (A)
3. (A) is true but (R) is false
4. Both (A) and (R) are false
View Answer
Water vapor \( H_2 O \) has a lower molar mass (approx. \(18 g/mol\) than nitrogen \( 28 g/mol\) and oxygen (\32 g/mol). Thus, replacing dry air molecules with \( H_2 O\) makes humid air less dense. Reason (R) is incorrect.
Assertion (A): Energy of molecules increase on increasing the temperature.
Reason (R): All substances expand on increasing the temperature.
1. (1) Both (A) & (R) are true and the (R) is the correct explanation of the (A)
2. (2) Both (A) & (R) are true but the (R) is not the correct explanation of the (A)
3. (3) (A) is true but (R) is false
4. (4) Both (A) and (R) are false
View Answer
Assertion (A) is true; average kinetic energy of molecules is directly proportional to absolute temperature. Reason (R) is false because some substances, like water between \(0^{circ}text{C}\) and \(4^{circ}text{C}\,) contract upon heating. Thus, (A) is true and (R) is false.
Assertion (A): During free expansion of an Ideal gas, entropy is zero.
Reason (R): Internal energy of an ideal gas is zero during free expansion.
1. (1) Both (A) & (R) are true and the (R) is the correct explanation of the (A)
2. (2) Both (A) & (R) are true but the (R) is not the correct explanation of the (A)
3. (3) (A) is true but (R) is false
4. (4) Both (A) and (R) are false
View Answer
During free expansion of an ideal gas, no work is done \(W=0\) and no heat is exchanged \(Q=0\). Therefore, change in internal energy \(Delta U = Q - W = 0\). For an ideal gas, \(Delta U = 0\) implies \(Delta T = 0\). Internal energy itself is not zero (it just doesn't change). Free expansion is an irreversible process, so entropy *increases* \(Delta S > 0\), it's not zero. Thus, both (A) and (R) are false.
Assertion (A): In an ideal monoatomic gas, The Internal energy of gas is equal to translational Kinetic energy of all its molecules
Reason (R): The Internal energy may get contributes from Translational, Rotatory, vibrationally as well as from the Potential energy corresponding to the molecular force.
1. (1) Both (A) & (R) are true and the (R) is the correct explanation of the (A)
2. (2) Both (A) & (R) are true but the (R) is not the correct explanation of the (A)
3. (3) (A) is true but (R) is false
4. (4) Both (A) and (R) are false
View Answer
Assertion (A) is true. For an ideal monatomic gas, molecules only have translational degrees of freedom, and there are no intermolecular forces, so internal energy consists solely of translational kinetic energy. Reason (R) is false; it describes contributions to internal energy from rotational, vibrational, and potential energies which are absent in an *ideal monatomic gas*.
Assertion (A): Energy of molecules increase on increasing the temperature.
Reason (R): All substances expand on increasing the temperature.
1. Both (A) & (R) are true and the (R) is the correct explanation of the (A)
2. Both (A) & (R) are true but the (R) is not the correct explanation of the (A)
3. (A) is true but (R) is false
4. Both (A) and (R) are false
View Answer
Concept: Temperature is a measure of average kinetic energy. Thermal expansion.
Formula: Average Kinetic Energy \( \propto T \).
Solution: Increasing temperature increases molecular kinetic energy. However, not all substances expand on heating (e.2.g., water between \( 0^{\circ}\text{C} \) and \( 4^{\circ}\text{C} \)). So R is false.
Assertion (A): During free expansion of an Ideal gas, entropy is zero.
Reason (R): Internal energy of an ideal gas is zero during free expansion.
1. Both (A) & (R) are true and the (R) is the correct explanation of the (A)
2. Both (A) & (R) are true but the (R) is not the correct explanation of the (A)
3. (A) is true but (R) is false
4. Both (A) and (R) are false
View Answer
Concept: Free expansion of an ideal gas. Entropy change. Internal energy.
Formula: For ideal gas, \( \Delta U = 0 \) (as \( Q=0, W=0 \)). Entropy change \( \Delta S > 0 \) for irreversible free expansion.
Solution: During free expansion of an ideal gas, \( \Delta U = 0 \) (meaning \( T \) is constant), but the internal energy itself is not zero. Also, free expansion is irreversible, so entropy *increases* (not zero). Both A and R are false.
Assertion (A): In an ideal monoatomic gas, The Internal energy of gas is equal to translational Kinetic energy of all its molecules
Reason (R): The Internal energy may get contributes from Translational, Rotatory, vibrationally as well as from the Potential energy corresponding to the molecular force.
1. Both (A) & (R) are true and the (R) is the correct explanation of the (A)
2. Both (A) & (R) are true but the (R) is not the correct explanation of the (A)
3. (A) is true but (R) is false
4. Both (A) and (R) are false
View Answer
Concept: Internal energy components for different types of gases.
Formula: For monoatomic ideal gas, \( U = \frac{3}{2} nRT \) (translational only).
Solution: For an ideal monoatomic gas, internal energy is purely translational kinetic energy. General internal energy can have translational, rotational, vibrational, and potential contributions (for real/complex gases), but potential energy is zero for ideal gases. A is true, R is true but not an explanation for A.