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NEET • PHYSICS • THERMODYNAMICS

Thermodynamics practice · Medium

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Showing 91-120 of 126 questions
91 Medium
91 of 126
Understanding +4 | -1
Compressed air in the tube of a wheel of a cycle at normal temperature suddenly starts coming out from a puncture. The air inside
View answer and explanation
Correct option: C
Starts becoming cooler
Option C (Starts becoming cooler) is correct because the sudden expansion of air is an adiabatic process, where the air does work on the surroundings, causing its internal energy and temperature to decrease.
92 Medium
92 of 126
Application +4 | -1
During an adiabatic expansion of 2 moles of a gas, the change in internal energy was found −50J- 50J.The work done during the process is
View answer and explanation
Correct option: D
50J50J
Option D (50J50J) is correct because in an adiabatic process, the work done is equal to the negative of the change in internal energy.
93 Medium
93 of 126
Analyzing +4 | -1
An engineer claims to have made an engine delivering 10kW10kW power with fuel consumption of 1g/s1g/s. The calorific value of the fuel is 2kcal/g2kcal/g. Is the claim of the engineer
View answer and explanation
Correct option: B
Invalid
Option B (Invalid) is correct because the claimed efficiency exceeds the maximum possible efficiency based on the fuel's calorific value.
94 Medium
94 of 126
Analyzing +4 | -1
Statement-1: When you touch two bodies, the body which is felt warmer must be of higher temperature than the other one.
Statement-2: The heat flows from higher temperature to lower temperature.
View answer and explanation
Correct option: B
Statement-1 is true, Statement-2 is true, Statement-2 is not the correct explanation for Statement-1
Option B (Statement-1 is true, Statement-2 is true, Statement-2 is not the correct explanation for Statement-1) is correct because while both statements are true, the perceived warmth is also influenced by the rate of heat transfer, not just the temperature difference.
95 Medium
95 of 126
Remembering +4 | -1
In an isothermal change, an ideal gas obeys
View answer and explanation
Correct option: A
Boyle’s law
Option A (Boyle’s law) is correct because Boyle's law states that for a fixed amount of gas at constant temperature, pressure and volume are inversely proportional (PV = constant), which is the condition for an isothermal change.
96 Medium
96 of 126
Application +4 | -1
For a gas, the difference between the two principle specific heats is 4150 Jkg−1K−1\mathrm{Jk}{\mathrm{g}^{ - 1}}{\mathrm{K}^{ - 1}}.What is the specific heat of the gas at constant volume if the ratio of specific heat is 1.4?
View answer and explanation
Correct option: B
10375Jkg−1K−110375\mathrm{Jk}{\mathrm{g}^{ - 1}}{\mathrm{K}^{ - 1}}
Option A (5186Jkg−1K−15186\mathrm{Jk}{\mathrm{g}^{ - 1}}{\mathrm{K}^{ - 1}}) is incorrect. Option B (10375Jkg−1K−110375\mathrm{Jk}{\mathrm{g}^{ - 1}}{\mathrm{K}^{ - 1}}) is the correct answer because it is derived from the given difference between specific heats and their ratio.
97 Medium
97 of 126
Understanding +4 | -1
Which of the following is not a thermodynamical function
View answer and explanation
Correct option: B
Work done
Option B (Work done) is correct because work done is a path function, not a state function like enthalpy, Gibb's energy, and internal energy.
98 Medium
98 of 126
Analyzing +4 | -1
An ideal monoatomic gas A\mathrm{A} is supplied heat so as to expand without changing its temperature. In another process, starting with same state, it is supplied heat at constant pressure. In both cases a graph of work done by gas is plotted versus heat added (Q) to the gas. The ratio of slope of graphs obtained in first & second process in η1\eta_1. The same ratio obtained for ideal diatomic gas is η2\eta_2. If the ratio η1η2=ab\frac{\eta_1}{\eta_2}=\frac{a}{b}. Then find a+ba+b.
View answer and explanation
Correct option: D
12
Option D (12) is correct. The ratio η1η2\frac{{\eta_1}}{{\eta_2}} is calculated based on the work done and heat added in isothermal and isobaric processes for monoatomic and diatomic gases.
99 Medium
99 of 126
Understanding +4 | -1
Assertion: The zeroth law said that, when two systems XX and YY are in thermal equilibrium, there must be a physical quantity that has the same value for both.
Reason: The physical quantity, which is same for both the systems, is temperature.
View answer and explanation
Correct option: A
If both assertion and reason are true and reason is the correct explanation of assertion
The zeroth law of thermodynamics defines thermal equilibrium, and temperature is the physical quantity that is the same for systems in thermal equilibrium. Therefore, both the assertion and reason are true, and the reason is the correct explanation.
100 Medium
100 of 126
Remembering +4 | -1
The specific heat relation for ideal gas is
View answer and explanation
Correct option: B
CP−CV=R{C_P} - {C_V} = R
Option B (CP−CV=R{C_P} - {C_V} = R) is correct because it is the correct specific heat relation for an ideal gas, where R is the gas constant.
101 Medium
101 of 126
Analyzing +4 | -1
Assertion: The ratio CPCV\frac{C_P}{C_V} is more for helium gas than for hydrogen gas.
Reason: Atomic mass of helium is more than that of hydrogen.
View answer and explanation
Correct option: B
If both Assertion & Reason are True but Reason is not a correct explanation of the Assertion
Option B is correct. The ratio CPCV\frac{C_P}{C_V} is indeed higher for helium than for hydrogen because helium is monoatomic (γ=5/3\gamma = 5/3) and hydrogen is diatomic (γ=7/5\gamma = 7/5). However, the reason given (atomic mass) is not the correct explanation; the correct reason is the difference in degrees of freedom.
102 Medium
102 of 126
Remembering +4 | -1
The mechanical equivalent of heat JJ is
View answer and explanation
Correct option: C
A conversion factor
Option C (A conversion factor) is correct because the mechanical equivalent of heat (J) is a factor that converts between units of energy (mechanical) and units of heat.
103 Medium
103 of 126
Application +4 | -1
500 J of heat energy is removed from 4 moles of a monoatomic ideal gas at constant volume. The temperature drops by
View answer and explanation
Correct option: C
10∘C{10^ \circ }C
Option C (10∘C{10^ \circ }C) is correct because the temperature drop is calculated using the heat removed, number of moles, and molar specific heat at constant volume.
104 Medium
104 of 126
Understanding +4 | -1
Assertion: Heat is not a state thermodynamic function.
Reason: Heat given in a system depends on process.
View answer and explanation
Correct option: A
If both Assertion & Reason are True & the Reason is a correct explanation of the Assertion
Heat is a path function, meaning its value depends on the process taken, not just the initial and final states. Therefore, both the assertion and reason are true, and the reason correctly explains the assertion.
105 Medium
105 of 126
Application +4 | -1
A Carnot’s engine works as a refrigerator between 250 K and 300 K. It receives 500 cal heat from the reservoir at the lower temperature. The amount of work done in each cycle to operate the refrigerator is
View answer and explanation
Correct option: A
420 J
Option A (420 J) is correct because it is the work done calculated from the Carnot refrigerator's coefficient of performance.
106 Medium
106 of 126
Understanding +4 | -1
Entropy of a system is related with
View answer and explanation
Correct option: A
Disorder of molecular motion of the system
Entropy is a measure of the disorder or randomness of a system, which is directly related to the disorder of molecular motion. Therefore, option A is correct.
107 Medium
107 of 126
Remembering +4 | -1
The ratio of the specific heats CpCV=γ\frac{C_p}{C_V}=\gamma in terms of degrees of freedom (n)(n) is given by
View answer and explanation
Correct option: C
(1+2n)\left(1+\frac{2}{n}\right)
Option C is correct because the relationship between the ratio of specific heats (γ\gamma) and the degrees of freedom (nn) is given by the formula γ=1+2n\gamma = 1 + \frac{2}{n}.
108 Medium
108 of 126
Understanding +4 | -1
The factor not needed to calculate heat lost or gained when there is no change of state is
View answer and explanation
Correct option: C
Relative density
Option C (Relative density) is correct because relative density is not directly used in the formula for calculating heat lost or gained without a change of state, which involves mass, specific heat, and temperature change.
109 Medium
109 of 126
Application +4 | -1
Calculate the time required to heat 20 kg20 \mathrm{~kg} of water from 10∘C10^{\circ} \mathrm{C} to 35∘C35^{\circ} \mathrm{C} using an immersion heater rated 1000 W1000 \mathrm{~W}. Assume that 80%80 \% of the power input is used to heat the water. Specific heat capacity of water =4200 J kg−1 K−1=4200 \mathrm{~J} \mathrm{~kg}^{-1} \mathrm{~K}^{-1}.
View answer and explanation
Correct option: B
44 min
Option B (44 min) is correct because it accurately calculates the time required based on the power, efficiency, mass, specific heat, and temperature change.
110 Medium
110 of 126
Understanding +4 | -1
Pressure – Temperature relationship for an ideal gas undergoing adiabatic change is (γ=CPCV)\left( {\gamma = \frac{{{C_P}}}{{{C_V}}}} \right)
View answer and explanation
Correct option: D
P1−γ  Tγ{P^{1 - \gamma }}\,\,{T^\gamma } = Constant
Option D (P1−γTγ{P^{1 - \gamma }}\\ {T^{\gamma }} = Constant) is correct because it represents the correct relationship between pressure and temperature in an adiabatic process.
111 Medium
111 of 126
Application +4 | -1
A gas under constant pressure of 4.5×1054.5 \times {10^5} Pa when subjected to 800 kJ of heat changes the volume from0.5m30.5{\mathrm{m}^3} to 2.0m32.0{\mathrm{m}^3}. The change in the internal energy of the gas is
View answer and explanation
Correct option: D
1.25×105J1.25 \times {10^5}\mathrm{J}
Option D (1.25×105J1.25 \times {10^5}\mathrm{J}) is correct because the change in internal energy is calculated using the first law of thermodynamics, accounting for the work done at constant pressure.
112 Medium
112 of 126
Understanding +4 | -1
The energy supply being cut-off, an electric heater element cools down to the temperature of its surroundings, but it will not cool further because
View answer and explanation
Correct option: D
Element & surroundings have same temp.
Option D (Element & surroundings have same temp.) is correct because heat transfer stops when thermal equilibrium is reached, meaning the element and surroundings are at the same temperature.
113 Medium
113 of 126
Analyzing +4 | -1
100 mol100 \mathrm{~mol} of an ideal gas is heated from 10∘10^{\circ} to 20∘C20^{\circ} \mathrm{C} keeping its (i) volume constant (ii) pressure constant. Let ΔUa\Delta U_a and ΔUb\Delta U_b denote the change in the internal energy of the gas due to process (i) and (ii), respectively. Then, which of the following shall be true?
View answer and explanation
Correct option: B
ΔUa>ΔUb\Delta U_a > \Delta U_b
Option A is correct because the internal energy change depends only on the temperature change for an ideal gas, and the temperature change is the same in both processes, but work done is different.
114 Medium
114 of 126
Remembering +4 | -1
The first operation involved in a carnot cycle is
View answer and explanation
Correct option: A
Isothermal Expansion
Option A (Isothermal Expansion) is correct because the Carnot cycle starts with isothermal expansion.
115 Medium
115 of 126
Analyzing +4 | -1
Heat energy of 184 kJ184 \mathrm{~kJ} is given to ice of mass 600 g600 \mathrm{~g} at −12∘C-12^{\circ} \mathrm{C}. Specific heat of ice is 2222.3 J kg−2222.3 \mathrm{~J} \mathrm{~kg}^{-} 10C−1{ }^{10} \mathrm{C}^{-1} and latent heat of ice in 336 kJ/kg−1336 \mathrm{~kJ} / \mathrm{kg}^{-1}

A. Final temperature of system will be 0∘C0^{\circ} \mathrm{C}.

B. Final temperature of the system will be greater than 0∘C0^{\circ} \mathrm{C}.

C. The final system will have a mixture of ice and water in the ratio of 5:15: 1.

D. The final system will have a mixture of ice and water in the ratio of 1:51: 5.

E. The final system will have water only.

Choose the correct answer from the options given below:
View answer and explanation
Correct option: B
A and D only
Options A and D are correct because the heat energy is sufficient to raise the ice to 0°C and partially melt it, resulting in a mixture of ice and water in a 5:1 ratio.
116 Medium
116 of 126
Remembering +4 | -1
The ideal gas equation for an adiabatic process is:
View answer and explanation
Correct option: A
PVγ=P V^\gamma= constant
Option A (PVγ=P V^{\gamma}= constant) is correct because it represents the relationship between pressure and volume in an adiabatic process.
117 Medium
117 of 126
Analyzing +4 | -1
CvandCp{C_v}and{C_p} denote the molar specific heat capacities of a gas at constant volume and constant pressure, respectively. Then
View answer and explanation
Correct option: B
Cp+Cv{C_p} + {C_v} is larger for a diatomic ideal gas than for a monoatomic ideal gas
Option B (Cp+Cv{C_p} + {C_v} is larger for a diatomic ideal gas than for a monoatomic ideal gas) is correct because both CpC_p and CvC_v are larger for diatomic gases due to their additional degrees of freedom.
118 Medium
118 of 126
Application +4 | -1
How much work to be done in decreasing the volume of an ideal gas by an amount of 2.4×10−4m32.4 \times {10^{ - 4}}{m^3} at normal temperature and constant normal pressure of 1×105N/m21 \times {10^5}N/{m^2}
View answer and explanation
Correct option: D
24J24J
Option D (24J24J) is correct because it is the work done calculated as W=PΔVW = P \Delta V at constant pressure.
119 Medium
119 of 126
Analyzing +4 | -1
Two Carnot engines A and B are operated in series. The engine AA receives heat from the source at temperature T1T_1 and rejects the heat to the sink at temperature T\mathrm{T}. The second engine BB receives the heat at temperature TT and rejects to its sink at temperature T2T_2. For what value of TT the efficiencies of the two engines are equal?
View answer and explanation
Correct option: D
T1T2\sqrt{T_1 T_2}
Option D is correct because for the efficiencies of the two Carnot engines to be equal, the intermediate temperature T must be the geometric mean of the source and sink temperatures.
120 Medium
120 of 126
Analyzing +4 | -1
An ideal monoatomic gas undergoes process PV1.25=P V^{1.25}= Constant . Then
View answer and explanation
Correct option: D
Heat is absorbed by gas if volume increases
Option D (Heat is absorbed by gas if volume increases) is correct because for the process PV1.25=ConstantPV^{1.25} = \text{Constant}, if volume increases, temperature decreases, but the work done is greater than the decrease in internal energy, leading to heat absorption.
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