Average translational kinetic energy per molecule is \(\frac{3}{2}kT\), which depends only on temperature and is same for all gases. However, at a given temperature, gas molecules have a distribution of speeds, not the same speed. Assertion is true but Reason is false.

Work and heat depend on the path, whereas internal energy depends only on the initial and final states. For any gas, \(C_P > C_V\) due to expansion work. Since volume is constant, \(dV = 0\), so work done \(W = 0\). Hence, all statements are correct.

Using the First Law of Thermodynamics, \(Delta Q = Delta U + W\). Here, \(Delta Q = 10\text{ J}\) and \(Delta U = -5\text{ J}\) (decrease). Substituting these values, \(10 = -5 + W\), which gives \(W = 15\text{ J}\).

The molar specific heat of a mixture at constant volume is \(C_{v,\text{mix}} = \frac{n_1 C_{v1} + n_2 C_{v2}}{n_1 + n_2}\). For monatomic helium, \(C_{v1} = 1.5R\), and for rigid diatomic hydrogen, \(C_{v2} = 2.5R\). Substituting gives \(C_{v,\text{mix}} = \frac{2(1.5R) + 3(2.5R)}{5} = 2.1R\).

Temperature gradient is defined as \(\frac{T_{\text{hot}} - T_{\text{cold}}}{L}\). Substituting the values, \(80 = \frac{30 - T_{\text{cold}}}{0.5}\). This simplifies to \(30 - T_{\text{cold}} = 40\), which gives \(T_{\text{cold}} = -10^\circ\text{C}\).

Velocity of sound is given by \(v = \sqrt{\frac{\gamma RT}{M}}\), so it depends on molar mass \(M\). Mechanical waves require a material medium. Humid air has lower density than dry air, which increases the speed of sound, making Statement C incorrect.

By kinetic theory: \(v_{\text{rms}} = \sqrt{\frac{3RT}{M}}\) -> (A)-(Q); Pressure \(P = \frac{1}{3} nm\bar{v}^2\) -> (B)-(P); Average KE \(= \frac{3}{2}k_BT\) -> (C)-(S); and Internal energy for diatomic gas \(= \frac{5}{2}RT\) -> (D)-(R).

According to Newton's law of cooling, \(\frac{T_1 - T_2}{\Delta t} = K\left(\frac{T_1+T_2}{2} - T_0\right)\). For the first interval, \(\frac{10}{t} = 65K\). For the second interval, \(\frac{20}{t'} = 50K\). Dividing these equations yields \(t' = \frac{13}{5}t\).

Total K.E. per molecule is \(\frac{f}{2}k_B T\). For monoatomic Argon, \(f=3\), and for diatomic Oxygen, \(f=5\). At equal temperature, the ratio of K.E. is \(3:5\).

Heat gained by water = \(100 \times 1 \times (90-30) = 6000\) cal. Heat lost by steam = \(m \times 540 + m \times 1 \times (100-90) = 550m\). Equating gives \(m \approx 11\) g.