Question 121:
easyIf \(10\text{ J}\) 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 \(W = 15\text{ J}\).
Question 122:
easyTwo 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 \(C_{v,\text{mix}} = \frac{2(1.5R) + 3(2.5R)}{5} = 2.1R\).
Question 123:
easyThe 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 \(T_{\text{cold}} = -10^\circ\text{C}\).
Question 124:
easyConsider the following statements.
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.
Question 125:
easyMatch 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).
Question 126:
easyA 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 \(t' = \frac{13}{5}t\).
Question 127:
easyThe 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.
Question 128:
easyThe internal energy of an ideal monoatomic gas increases by the same amount as work done on the gas, then:
By the first law, \(dQ = dU + dW\). If work is done on the gas, \(dW_{\text{by}} = -dW_{\text{on}}\). Given \(dU = dW_{\text{on}}\), so \(dQ = 0\). This represents an adiabatic process.
Question 129:
easyThe work done by 3 moles of gas at 47°C to triple its volume at constant pressure is (\(R = 2\) cal mol\(^{-1}\) °C\(^{-1}\)):
At constant pressure, \(W = nR\Delta T\). Since volume triples, temperature also triples (from \(320\) K to \(960\) K), so \(\Delta T = 640\) K. Work \(W = 3 \times 2 \times 640 = 3840\) cal.
Question 130:
easyThe 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 \(3:5\).