Assertion (A):The average translational kinetic energy per molecule of gas for various gases at the same temperature is the same.
Reason (R):At a given temperature, all molecules of a gas move with nearly the same speed.
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.
A. Work and heat are path functions in thermodynamics.
B. The internal energy of a gaseous system is state function.
C. For gaseous system, \(C_P\) is greater than \(C_V\).
D. Work done by gas at constant volume is zero.
Based on above information pick the correct option.
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.
If \(10\text{ J}\) of heat energy is supplied to a gas sample and \(5\text{ J}\) of its internal energy decreases during the process, then work done by the gas will be
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 \(W = 15\text{ J}\).
Two moles of helium gas is mixed with three moles of hydrogen gas (taken to be rigid). The molar specific heat of mixture at constant volume will be
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 \(C_{v,\text{mix}} = \frac{2(1.5R) + 3(2.5R)}{5} = 2.1R\).
The temperature gradient in a rod \(0.5\text{ m}\) long is \(80^\circ\text{C/m}\). If temperature of hotter end is \(30^\circ\text{C}\), then temperature of the colder end will be
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 \(T_{\text{cold}} = -10^\circ\text{C}\).
Statement A: Velocity of sound in gaseous medium depends on molar mass of gas.
Statement B: Mechanical wave require a material medium for their propagation.
Statement C: Speed of sound is less in humid air.
Which of the statement(s) is/are correct?
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.
Match Column – I and Column – II and choose the correct match from the given choices.
Column-I
(A) Root mean square speed of gas molecules
(B) Pressure exerted by ideal gas
(C) Average kinetic energy of a molecule
(D) Total internal energy of 1 mole of a diatomic gas
Column-II
(P) \(\frac{1}{3} n m \bar{v}^2\)
(Q) \(\sqrt{\frac{3RT}{M}}\)
(R) \(\frac{5}{2} RT\)
(S) \(\frac{3}{2} k_B T\)
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).
A cup of coffee cools from \(90^\circ\text{C}\) to \(80^\circ\text{C}\) in \(t\) minutes, when the room temperature is \(20^\circ\text{C}\). The time taken by a similar cup of coffee to cool from \(80^\circ\text{C}\) to \(60^\circ\text{C}\) at a room temperature same at \(20^\circ\text{C}\) is
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 \(t' = \frac{13}{5}t\).
The ratio of total K.E of a molecule of Argon and Oxygen gas at 27°C is equal to:
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 \(3:5\).
The temperature of 100 g of water is to be raised from 30 °C to 90 °C by adding steam to it. The mass of steam required to raise this temperature will be nearly (Take, \(S_W = 1\) cal/g °C, \(L = 540\) cal/g):
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.