Assertion (A): The ratio \( \frac{C_P}{C_V} \) is more for helium gas than for hydrogen gas.
Reason (R): Atomic mass of helium is more than that of hydrogen.
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
For Helium (monoatomic), \( \gamma = 5/3 \). For Hydrogen (diatomic), \( \gamma = 7/5 \). Since \( 5/3 > 7/5 \), (A) is true. Atomic mass of He is 4 amu, H is 1 amu (H2 is 2 amu), so (R) is true.
However, \( \gamma \) depends on degrees of freedom (monoatomic vs diatomic), not atomic mass. So, (R) is not the correct explanation.
Assertion (A): On a V-T graph, the slope of an isobar increases with pressure.
Reason (R): At constant temperature, for an ideal gas its volume is directly proportional to its pressure.
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
For an isobar, \( V = (\frac{nR}{P})T \). The slope on a V-T graph is \( \frac{nR}{P} \). As P increases, slope decreases, so (A) is false. Boyle's law states that at constant T, \( V \propto \frac{1}{P} \), i.e., V is inversely proportional to P, so (R) is false.
Assertion (A): Internal energy of real gas is always negative at absolute zero temperature.
Reason (R): Potential energy of a bounded system is negative.
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
At absolute zero, kinetic energy is minimal (zero for ideal gas). For a real gas, attractive intermolecular forces mean potential energy is negative (relative to infinite separation). So, total internal energy is negative. Thus, (A) is true. (R) is also true, as attractive forces in a bounded system lead to negative potential energy. (R) explains (A).
Assertion (A): The average translational kinetic energy of the molecules in one mole of all ideal gases, at the same temperature is the same.
Reason (R): The average kinetic energy of one mole of any ideal gas at temperature T is given by \( \frac{3}{2}RT \).
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
The average translational kinetic energy per mole for any ideal gas is \( \frac{3}{2}RT \), dependent only on T.
So (A) is true. The formula in (R) represents this average translational kinetic energy per mole. So (R) is true and correctly explains (A).
Assertion (A): For an ideal gas, at constant temperature, the product of the pressure and volume is constant.
Reason (R): The mean square velocity of gas molecules is inversely proportional to mass of molecule.
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
Boyle's Law states that for an ideal gas at constant T, \( PV = \text{constant} \). So (A) is true. The mean square velocity \( = \frac{3kT}{m} \), so it is inversely proportional to molecular mass m.
So (R) is true. However, (R) does not explain Boyle's law (A).
A container of volume \(200 \text{cm}^3\) contains 0.2 mole of hydrogen gas and 0.3 mole of argon gas. The pressure of the system at temperature 200 K (\(R = 8.3 \text{J} \text{K}^{-1} \text{mol}^{-1}\)) will be
1. \(4.15 \times 10^5 \text{Pa}\)
2. \(4.15 \times 10^6 \text{Pa}\)
3. \(6.15 \times 10^5 \text{Pa}\)
4. \(6.15 \times 10^4 \text{Pa}\)
View Answer
Total moles \(n = 0.2 + 0.3 = 0.5\). Using \(P = \frac{nRT}{V} = \frac{0.5 \times 8.3 \times 200}{200 \times 10^{-6}} = 4.15 \times 10^6 \text{Pa}\).
Four particles have velocities 1, 0, 2 and 3 m/s. The root mean square velocity of the particles is
1. \(\sqrt{3.5} \text{m/s}\)
2. \(\sqrt{5.3} \text{m/s}\)
3. \(\sqrt{2.8} \text{m/s}\)
4. \(\sqrt{4.7} \text{m/s}\)
View Answer
The RMS velocity is given by \[v_{\text{rms}} = \sqrt{\frac{v_1^2 + v_2^2 + v_3^2 + v_4^2}{4}}\]. Substituting the values, \[v_{text{rms}} = \sqrt{\frac{1^2 + 0^2 + 2^2 + 3^2}{4}} = \sqrt{\frac{14}{4}} = \sqrt{3.5} \text{m/s}\].
Consider the following statements:
A. The temperature at which the liquid and solid state of substance is in thermal equilibrium with each other is called its melting point.
B. The temperature at which the liquid and the vapour states of the substance coexist is called its boiling point.
C. The change from solid state to vapour state without passing through liquid state is called sublimation.
Based on above information pick correct option.
1. Only A and B correct
2. Only C and B are correct
3. Only C is incorrect
4. All statements A, B and C are correct
View Answer
All statements correctly define physical principles: statement A defines melting point, statement B defines boiling point, and statement C defines sublimation.
Match list-I with list-II about effect of various factors on speed of sound in a gas.
\[\begin{array}{|l|l|} \hline
\text{List-I (Factor)} & \text{List-II (Speed)} \\ \hline
\text{a. With increase in temperature} & \text{(ii) Increases} \\ \hline
\text{b. With increase in molecular weight of gas (temp const)} & \text{(iii) Decreases} \\ \hline
\text{c. With increase in pressure at constant temperature} & \text{(i) Same} \\ \hline
\end{array}\]
1. a(ii), b(i), c(iii)
2. a(ii), b(iii), c(i)
3. a(iii), b(ii), c(i)
4. a(i), b(ii), c(ii)
View Answer
Speed of sound is \(v = \sqrt{\frac{\gamma RT}{M}}\). Temperature increase increases speed, molecular weight increase decreases speed, and pressure has no effect at constant temperature.
In a thermodynamic process, pressure (in Pa) varies as, \(P = a + bV\) [where \(V\) represents volume in \(\text{m}^3\), \(a\) and \(b\) are constants]. The work done by gas during expansion from \(2 \text{m}^3\) to \(3 \text{m}^3\) will be
1. \(\left(\frac{3a}{2} + b \right) \text{J}\)
2. \(\left(a + \frac{5b}{2}\right) \text{J}\)
3. \(\left(b + \frac{5a}{2}\right) \text{J}\)
4. Zero
View Answer
Work done \(W = \int_{V_1}^{V_2} P dV = \int_2^3 (a+bV) dV = \left[aV + \frac{bV^2}{2}\right]_2^3 = a(3-2) + \frac{b}{2}(9-4) = a + \frac{5b}{2}\).