Question 281:
easyStatement I: Internal energy of an ideal gas remains constant in an adiabatic process.
Statement II: In an adiabatic process, change in internal energy of a gas is equal to work done on or by the gas in the process.
In an adiabatic process, \(Q = 0\), so \(\Delta U = -W\), which means internal energy changes, so Statement I is incorrect. Statement II is correct since change in internal energy corresponds directly to the work done on or by the gas.
Question 282:
easyTemperature of a body rises by \(2^{\circ}C\), the corresponding temperature rise in Kelvin will be
The change in temperature on the Celsius scale is equal to the change in temperature on the Kelvin scale because the size of one degree Celsius is equal to one Kelvin: \(\Delta T_{text{C}} = \Delta T_{\text{K}} = 2\text{ K}\).
Question 283:
easyThe physical quantity that determines whether or not the given system A is in thermal equilibrium with another system B is called
According to the Zeroth Law of Thermodynamics, temperature is the physical quantity that determines if systems are in thermal equilibrium with each other.
Question 284:
moderateA body cools from \(80^\circ\text{ C}\) to \(70^\circ\text{ C}\) in \(12\text{minutes}\) and from \(70^\circ\text{ C}\) to \(60^\circ\text{ C}\) in \(t \text{ minutes}\). The value of \(t\) is (temperature of surrounding is \(40^\circ\text{ C}\).
According to Newton's Law of Cooling, \(\frac{T_i - T_f}{t} = K\left(\frac{T_i + T_f}{2} - T_s\right)\). For the first interval, \(\frac{80-70}{12} = K(75-40) ⇒ K = \frac{1}{42}\). For the second interval, \(\frac{70-60}{t} = K(65-40) ⇒ \frac{10}{t} = \frac{25}{42}\), which gives \(t = 16.8\text{ minutes}\).
Question 285:
easyIf transmittance of a surface is \(\frac{1}{7}\), reflectance is \(\frac{1}{8}\), then the absorptance of the surface will be
By conservation of energy, the sum of absorptance \(a\), reflectance \(r\), and transmittance \(t\) is equal to \(1\): \(a + r + t = 1\). Therefore, \(a = 1 - \frac{1}{8} - \frac{1}{7} = 1 - \frac{15}{56} = \frac{41}{56}\).
Question 286:
moderateA solid cube of side \(4\text{ m}\) having coefficient of areal expansion \(2 \times 10^{-5}/^\circ\text{C}\). If temperature is changed by \(40^\circ\text{C}\) then the change in side length of the cube will be
Coefficient of linear expansion is \(\alpha = \frac{\beta}{2} = 1 \times 10^{-5}/^\circ\text{C}\). Change in length \(\Delta L = L \alpha \Delta T = 4 \times (1 \times 10^{-5}) \times 40 = 1.6 \times 10^{-3}\text{ m} = 1.6\text{ mm}\).
Question 287:
moderateThermal capacity of \(40\text{ g}\) of aluminium of specific heat \(0.2\text{ cal/(g }^\{circ}\text{C)}\) is
Thermal capacity is given by the formula \(C = m \cdot s\). Substituting the values, \(C = 40\text{ g} \times 0.2\text{ cal/(g }^\circ\text{C)} = 8\text{ cal/}^\circ\text{C}\.
Question 288:
easyThe physical quantity that determines whether or not the given system A is in thermal equilibrium with another system B is called
Temperature is the physical quantity that determines thermal equilibrium. Two systems are in thermal equilibrium if and only if they are at the same temperature.
Question 289:
easyOn increasing the number density for a gas in a vessel, mean free path of the gas will
The mean free path is given by \(\lambda = \frac{1}{\sqrt{2} n \pi d^2}\). Since \(\lambda\) is inversely proportional to the number density \(n\), increasing the number density decreases the mean free path.
Question 290:
Equal masses of an ideal gas are sealed in two vessels one of pressure \(P_0\) and other of pressure \(2P_0\). If first vessel is at temperature of 400 K and the other is at 600 K. Find the ratio of volume of two container.
From the ideal gas law \(PV = nRT\), the volume is proportional to \(\frac{T}{P}\) for equal masses of the same gas. Thus, \(\frac{V_1}{V_2} = \frac{T_1}{T_2} \times \frac{P_2}{P_1} = \frac{400}{600} \times \frac{2P_0}{P_0} = \frac{4}{3}\).