Question 311:
easyThe internal energy of an ideal gas depends upon
The internal energy of an ideal gas is a function of temperature only, as there are no intermolecular forces of attraction in an ideal gas. Therefore, \( U \propto T \).
Practice NEET Thermal Physics Questions
Question 312:
easyA monoatomic gas does 150 J of work in isothermal expansion. The heat supplied to the gas is
For an isothermal process, the change in internal energy is \( \Delta U = 0 \). According to the first law of thermodynamics, \( Q = \Delta U + W \), which gives \( Q = 0 + 150\text{ J} = 150\text{ J} \).
Question 313:
moderateIn thermodynamic processes, correct match of column-I with column-II is:
| Column-I (Type of process) | Column-II (Feature) |
| a. Isothermal | (iv) Temperature constant |
| b. Isobaric | (ii) Pressure constant |
| c. Isochoric | (i) Volume constant |
| d. Adiabatic | (iii) No heat flow between system and surroundings |
Isothermal process has constant temperature (a-iv). Isobaric has constant pressure (b-ii). Isochoric has constant volume (c-i). Adiabatic has no heat flow (d-iii). Matching these gives option D.
Question 314:
easyFor an ideal gas, total energy is equally distributed in all possible energy modes, with each mode has an average energy equal to \(\frac{1}{2} k_B T\), and each vibrational mode has energy contribution of
Each vibrational mode has both kinetic energy and potential energy modes, thus having two degrees of freedom. Therefore, the average energy per vibrational mode is \(2 \times \frac{1}{2} k_B T = k_B T\).
Question 315:
moderate\(15\text{ gm}\) of ice at \(0^\circ\text{C}\) is mixed with \(300\text{ gm}\) of water at \(50^\circ\text{C}\) in a container. There is no heat loss due to radiation and water equivalent of container is ignored. What will be final temperature of water?
Heat absorbed to melt ice: \(Q_1 = 15 \times 80 = 1200\text{ cal}\). Let final temperature be \(T\). Heat gained by melted ice: \(15 T\). Heat lost by hot water: \(300(50 - T)\). Equilibrium: \(1200 + 15T = 300(50-T) \implies 315T = 13800 \implies T \approx 43.8^\circ\text{C}\).
Question 316:
easyThe relation between coefficient of linear expansion (\(\alpha\)) and coefficient of volume expansion (\(\gamma\)) for solids is
Coefficient of volume expansion is three times the coefficient of linear expansion for an isotropic solid, i.e., \(\gamma = 3\alpha\).
Question 317:
moderateMercury thermometer can be used to measure temperature upto: (1992)
The boiling point of mercury is approximately $356.7^\circ\text{C}$. Therefore, a standard mercury thermometer can measure temperatures up to around $360^\circ\text{C}$.
Question 318:
moderateA Centigrade and a Fahrenheit thermometer are dipped in boiling water. The water temperature is lowered until the Fahrenheit thermometer registers $140^\circ\text{F}$. What is the fall in temperature as registered by the centigrade thermometer? (1990)
Initial temperature of boiling water is $212^\circ\text{F}$. Fall in Fahrenheit $= 212^\circ\text{F} - 140^\circ\text{F} = 72^\circ\text{F}$. Using $\Delta C = \frac{5}{9} \Delta F$, fall in Celsius $= \frac{5}{9} \times 72 = 40^\circ\text{C}$.
Question 319:
moderateThe value of coefficient of volume expansion of glycerin is $5 \times 10^{-4}\text{ /K}$. The fractional change in the density of glycerin for a rise of $40^\circ\text{C}$ in its temperature, is: (2015 Re)
The fractional change in density is approximately given by $\frac{\Delta \rho}{\rho} = \gamma \Delta T$. Substituting the values: $\frac{\Delta \rho}{\rho} = 5 \times 10^{-4} \times 40 = 200 \times 10^{-4} = 0.020$.
Question 320:
moderateWhich of the following rods, (given radius $r$ and length $l$) each made of the same material and whose ends are maintained at the same temperature will conduct most heat? (2005)
The rate of heat conduction is $H = \frac{KA\Delta T}{l} = \frac{K(\pi r^2)\Delta T}{l}$. Since material and $\Delta T$ are same, $H \propto \frac{r^2}{l}$. This ratio is maximum for $r = 2r_0$ and $l = l_0$ (ratio $\propto 4$).