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NEET • PHYSICS • THERMODYNAMICS

Thermodynamics practice · Mixed difficulty

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Showing 91-120 of 183 questions

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91 Medium
91 of 183
Analyzing +4 | -1
Statement-I: Monoatomic, diatomic and polyatomic gases are adiabatically compressed such that compression ratio is (P2P1)\left(\frac{P_2}{P_1}\right). Then monoatomic gas will have maximum final volume out of these three gases.
Statement-II: Monoatomic gas has least degree of freedom.
View answer and explanation
Correct option: A
Statement-I is true and statement-II is true
Statement-I is true because monoatomic gases have the highest adiabatic index, leading to the smallest volume change during adiabatic compression. Statement-II is true because monoatomic gases have the fewest degrees of freedom (3).
92 Medium
92 of 183
Application +4 | -1
252 micro coulombs of charge is transferred between two points having a potential difference of 109 volt during an electric discharge. What mass of ice (at 00C)\left( {at\,\mathrm{ }{0^0}C} \right) would it melt if all the energy released could be used for the above purpose? (take mechanical equivalent of heat as 4.2 joule/calorie and latent heat of fusion of ice is 80 cal/gm)
View answer and explanation
Correct option: B
0.75 kg
Option B (0.75 kg) is correct because the energy released during the electric discharge is used to melt the ice, and the mass of ice melted can be calculated using the latent heat of fusion.
93 Medium
93 of 183
Analyzing +4 | -1
(A): Air in a cylinder is suddenly compressed by a piston. If the volume after compression is half the original volume the final pressure is twice the initial pressure.
(R): Suddenly compressed process is isobaric process.
View answer and explanation
Correct option: D
Both (A) and (R) are false
Option D is correct because the compression is adiabatic, not isobaric, and the final pressure is not twice the initial pressure for an adiabatic process.
94 Medium
94 of 183
Analyzing +4 | -1
An engine operating between the boiling and freezing points of water will have
A. Efficiency more than 27%27 \%
B. Efficiency less than the efficiency of a Carnot engine operating between the same two temperatures.
C. Efficiency equal to 27%.27\%.
D. Efficiency less than 27%.27\%.
Choose the correct answer from the options given below
View answer and explanation
Correct option: C
B and D only
Option C (B and D only) is correct because any real engine operating between two temperatures will have efficiency less than the Carnot engine, and the Carnot efficiency between 0°C and 100°C is less than 27%.
95 Medium
95 of 183
Remembering +4 | -1
First law of thermodynamics is given by
View answer and explanation
Correct option: A
dQ=dU+PdVdQ = dU + PdV
Option A (dQ=dU+PdVdQ = dU + PdV) is correct because it represents the mathematical formulation of the first law of thermodynamics.
96 Medium
96 of 183
Understanding +4 | -1
Air in a cylinder is suddenly compressed by a piston, which is then maintained at the same position. With the passage of
View answer and explanation
Correct option: A
The pressure decreases
Option A (The pressure decreases) is correct because when air is suddenly compressed, the temperature increases. However, with the passage of time, the heat dissipates to the surroundings, and the temperature returns to the initial value. Since the volume is smaller, the pressure decreases to maintain thermal equilibrium.
97 Medium
97 of 183
Understanding +4 | -1
Which of the following can not determine the state of a thermodynamic system
View answer and explanation
Correct option: D
Any one of pressure, volume or temperature
Option D (Any one of pressure, volume or temperature) is correct because to define the state of a thermodynamic system, you need at least two independent state variables. Knowing only one is insufficient.
98 Medium
98 of 183
Application +4 | -1
Air is expanded form 50 litre to 150 litre at 2 atmospheric pressure (1 atm pressure = 105kgm−2\mathbf{1}{\mathbf{0}^\mathbf{5}}\mathbf{kg}\mathrm{ }{\mathbf{m}^{ - \mathbf{2}}}). The external work done is
View answer and explanation
Correct option: C
2×104J2 \times {10^4}J
The external work done is 2×104J2 \times {10^4}J because it is calculated using the formula W=PΔVW = P\Delta V, where P is the pressure and ΔV\Delta V is the change in volume.
99 Medium
99 of 183
Remembering +4 | -1
Which of the following law of thermodynamics forms the basis for the definition of temperature?
View answer and explanation
Correct option: C
Zeroth law
Option C (Zeroth law) is correct because the zeroth law of thermodynamics defines thermal equilibrium and forms the basis for temperature measurement.
100 Medium
100 of 183
Application +4 | -1
A diatomic gas (γ=1.4)\left( {\gamma \mathrm{ } = \mathrm{ }1.4} \right) does 60 J of work when it is expanded isobarically. The heat given to gas in the process is
View answer and explanation
Correct option: A
210 J
Option A (210 J) is correct because the heat given to the gas is calculated using the relationship between work done and heat in an isobaric process for a diatomic gas.
101 Medium
101 of 183
Application +4 | -1
In supplying 400 calories of heat to a system, the work done will be
View answer and explanation
Correct option: B
1672 joules
Option B (1672 joules) is correct because it is the heat in joules equivalent to 400 calories.
102 Easy
102 of 183
Application +4 | -1
Calculate the work done when one mole of a perfect gas is compressed adiabatically. The initial pressure and volume of the gas are 105 N/m210^5 \mathrm{~N} / \mathrm{m}^2 and 6 litre respectively. The final volume of the gas is 2 litre. Molar specific heat of the gas at constant volume is 3R2\frac{3 R}{2}.
View answer and explanation
Correct option: A
957
The correct answer is 957. The work done during an adiabatic compression is calculated using the formula W=P2V2−P1V11−γW = \frac{P_2V_2 - P_1V_1}{1-\gamma}, where γ=1+2f\gamma = 1 + \frac{2}{f} and f is the degrees of freedom.
103 Medium
103 of 183
Remembering +4 | -1
How much heat (in joules) is required to raise the temperature of 1 g1 \mathrm{~g} of water from 14.5∘C14.5^{\circ} \mathrm{C} to 15.5∘C15.5^{\circ} \mathrm{C} at 1 atmospheric pressure?
View answer and explanation
Correct option: B
4.186 J4.186 \mathrm{~J}
Option B (4.186 J4.186 \mathrm{~J}) is correct because it represents the specific heat capacity of water, which is the energy required to raise the temperature of 1 gram of water by 1 degree Celsius.
104 Medium
104 of 183
Application +4 | -1
A lead bullet of 10 g travelling at 300 m/s300\,m/s strikes against a block of wood and comes to rest. Assuming 50%50\% of heat is absorbed by the bullet, the increase in its temperature is (specific heat of lead =150J/kg,K)= 150J/kg,K)
View answer and explanation
Correct option: C
1500C{150^0}C
The kinetic energy of the bullet is converted into heat. Half of this heat is absorbed by the bullet, causing its temperature to rise. Using the formula Q=mcΔTQ = mc\Delta T, we find the temperature increase to be 150°C.
105 Medium
105 of 183
Application +4 | -1
Latent heat of 1gm of steam is 536 cal/gm, then its value in joule/kg is
View answer and explanation
Correct option: A
2.25×1062.25 \times {10^6}
Option A (2.25×1062.25 \times {10^6}) is correct because it represents the latent heat of steam converted from cal/gm to joule/kg using the appropriate conversion factor.
106 Medium
106 of 183
Understanding +4 | -1
Statement-1: The quantity of heat in a solid body determines its temperature.
Statement-2: Whenever heat is supplied to a solid body which is not undergoing phase change the temperature of the solid body increases.
View answer and explanation
Correct option: D
Statement-1 is false, Statement-2 is true
Statement-1 is false because temperature is a measure of the average kinetic energy of the molecules, not the quantity of heat. Statement-2 is true because adding heat to a solid generally increases its temperature unless a phase change is occurring.
107 Medium
107 of 183
Analyzing +4 | -1
Assertion: P{P} versus 1V\frac{1}{{V}} graph is straight line for adiabatic process
Reason: PV=PV = constant for adiabatic process.
View answer and explanation
Correct option: D
If both Assertion and Reason are incorrect
Option D (If both Assertion and Reason are incorrect) is correct because the P vs 1/V graph for an adiabatic process is NOT a straight line, and PV = constant is only true for an isothermal process, not adiabatic.
108 Easy
108 of 183
Application +4 | -1
An ideal gas has a molar heat capacity C=9R2\mathrm{C}=\frac{9 \mathrm{R}}{2} in the process PT=\mathrm{PT}= constant. Find the degree of freedom of the gas molecule. (R=(\mathrm{R}= ideal gas constant)
View answer and explanation
Correct option: A
5
Option A (5) is correct because the molar heat capacity is related to the degrees of freedom by the equation C = Cv + R/2, and Cv = (f/2)R.
109 Medium
109 of 183
Application +4 | -1
A Carnot engine working between the temperature 500 K and 1000 K does a work of 1000 J. The heat energy supplied by the source to the engine per cycle is
View answer and explanation
Correct option: C
2000 J
Option C (2000 J) is correct because it is the heat energy supplied calculated from the Carnot engine efficiency and work done.
110 Easy
110 of 183
Understanding +4 | -1
A gas has nn degrees of freedom. The ratio of specific heat of gas at constant volume to the specific heat of gas at constant pressure will be
View answer and explanation
Correct option: A
nn+2\frac{n}{n+2}
Option A (nn+2\frac{n}{n+2}) is correct because the ratio of specific heat at constant volume (CvC_v) to specific heat at constant pressure (CpC_p) is given by CvCp=1γ=n/2n/2+1=nn+2\frac{C_v}{C_p} = \frac{1}{\gamma} = \frac{n/2}{n/2 + 1} = \frac{n}{n+2}, where n is the degrees of freedom.
111 Easy
111 of 183
Understanding +4 | -1
The work done in an isothermal expansion of a gas depends upon
View answer and explanation
Correct option: C
Both temperature and expansion ratio
Option C is correct because the work done in an isothermal expansion of a gas depends on both the temperature and the expansion ratio (ratio of final to initial volume).
112 Hard
112 of 183
Application +4 | -1
NN moles of non-linear polyatomic gas (degree of freedom 6) is mixed with 2 moles of monoatomic gas. The resultant mixture has molar specific heat equal to that of a diatomic gas, then NN is
View answer and explanation
Correct option: D
4
Option D (4) is correct because the number of moles N is calculated by equating the molar specific heat of the mixture to that of a diatomic gas.
113 Medium
113 of 183
Remembering +4 | -1
Which of the following the scalar quantity is
View answer and explanation
Correct option: A
Temperature
Option A (Temperature) is correct because temperature is a scalar quantity, defined only by magnitude.
114 Medium
114 of 183
Understanding +4 | -1
In an isothermal change of an ideal gas, ΔU=0\Delta U = 0. The change in heat energy ΔQ\Delta Q is equal to
View answer and explanation
Correct option: B
WW
Option B (WW) is correct because in an isothermal process with an ideal gas, the change in internal energy is zero, so the heat energy change equals the work done.
115 Medium
115 of 183
Analyzing +4 | -1
Assertion: first law of thermodynamics allows many processes which actually don't happen.
Reason: First law of thermodynamics must not be violated for any process to happen.
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 (If both Assertion & Reason are True but Reason is not a correct explanation of the Assertion) is correct because the first law doesn't forbid processes that violate the second law, and the reason is true but doesn't explain the assertion.
116 Medium
116 of 183
Analyzing +4 | -1
Given below are two statements: one is labelled as Assertion A and the other is labelled as Reason R

Assertion A: Efficiency of a reversible heat engine will be highest at −273∘C- 273^\circ C temperature of cold reservoir.

Reason R: The efficiency of Carnot’s engine depends not only on temperature of cold reservoir but it depends on the temperature of hot reservoir too and is given as η=(1−T2T1)\eta = \left( {1 - \frac{{{T_2}}}{{{T_1}}}} \right).
In the light of the above statements, choose the correct answer from the options given below
View answer and explanation
Correct option: A
Both A and R are true and R is the correct
explanation of A
The correct answer is A because the efficiency of a reversible heat engine is highest at the lowest cold reservoir temperature, and the Carnot efficiency formula explicitly shows dependence on both hot and cold reservoir temperatures.
117 Easy
117 of 183
Understanding +4 | -1
Statement-1: A quasi-static process is so called because it is a sudden and large change of the system.
Statement-2: An adiabatic process is no quasi-static because it is a sudden and large change of the system.
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 D is correct. Statement-1 is false because a quasi-static process is a slow, gradual change. Statement-2 is true because an adiabatic process can be a sudden change, and therefore not quasi-static.
118 Hard
118 of 183
Application +4 | -1
Work done in rising temperature of five mole of Helium adiabatically through θ∘C\theta^{\circ} {C} is W{W}. Then the work done in an isobaric process of the sample for same change of temperature is
View answer and explanation
Correct option: D
−2W3-\frac{2 W}{3}
Option D (−2W3-\frac{2 W}{3}) is correct because the ratio of work done in the isobaric process to the adiabatic process is -2/3 based on the specific heat capacities.
119 Easy
119 of 183
Application +4 | -1
Two moles a monoatomic gas is mixed with six moles of a diatomic gas. The molar specific heat of the mixture at constant volume is:
View answer and explanation
Correct option: B
94R\frac{9}{4} R
Option B (94R\frac{9}{4} R) is correct because the molar specific heat of the mixture is calculated using the weighted average of the individual molar specific heats.
120 Hard
120 of 183
Application +4 | -1
A diatomic gas (γ=1.4)(\gamma=1.4) does 100 J100 \mathrm{~J} of work in an isobaric expansion. The heat given to the gas is:
View answer and explanation
Correct option: A
350 J350 \mathrm{~J}
Option A (350 J350 \mathrm{~J}) is correct because it is calculated using the relationship between heat, work, and the adiabatic index for an isobaric process.
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