31 Year NEET Previous Year Questions: Thermodynamics - 1
14 Questions MCQ Test Physics Class 11 | 31 Year NEET Previous Year Questions: Thermodynamics - 1
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In thermodynamic processes which of the following statements is not true? [2009]
In an isochoric process volume remains constant whereas pressure remains constant in isobaric process.
The internal energy change in a system that has absorbed 2 kcals of heat and done 500 J of work is: [2009]
According to first law of thermodynamics
Q = ΔU + W
ΔU = Q – W
= 2 × 4.2 × 1000 – 500 = 8400 –500
= 7900 J
If ΔU and ΔW represent the increase in internal energy and work done by the system respectively in a thermodynamical process, which of the following is true? [2010]
By first law of thermodynamics,
ΔQ = ΔU + ΔW
In adiabatic process, ΔQ = 0
∴ ΔU = -ΔW
In isothermal process, ΔU = 0
∴ ΔQ -ΔW
During an isothermal expansion, a confined ideal gas does –150 J of work against its surroundings. This implies that [2011]
From the first law of thermodynamics
ΔU = Q + W
For isothermal process, ΔU = 0
Therefore, Q = - W
Given W = - 150
Therefore, Q = + 150
When Q is positive, the heat is added to the gas.
When 1 kg of ice at 0°C melts to water at 0°C, the resulting change in its entropy, taking latent heat of ice to be 80 cal/°C, is [2011]
Change in entropy is given by
A mass of diatomic gas (γ= 1.4) at a pressure of 2 atmospheres is compressed adiabatically so that its temperature rises from 27°C to 927°C.The pressure of the gas in final state is [2011M]
T1 = 273 + 27 = 300K
T2 = 273 + 927 = 1200K
For adiabatic process,
= P1 (27) = 2 × 128 = 256 atm
A thermodynamic system is taken through the cycle ABCD as shown in figure. Heat rejected by the gas during the cycle is : [2012]
∵ Internal energy is the state function.
∴ In cyclie process; ΔU = 0
According to 1st law of thermodynamics
So heat absorbed ΔQ = W = Area under the curve
= – (2V) (P) = – 2PV
So heat rejected = 2PV
One mole of an ideal gas goes from an initial state A to final state B via two processes : It first undergoes isothermal expansion from volume V to 3V and then its volume is reduced from 3V to V at constant pressure. The correct P-V diagram representing the two processes is : [2012]
1st process is isothermal expansion which is only correct shown in option (d) 2nd process is isobaric compression which is correctly shown in option (d)
If Q1, Q2, Q3 indicate the heat a absorbed by the gas along the three processes and ΔU1, ΔU2, ΔU3 indicate the change in internal energy along the three processes respectively, then
Initial and final condition is same for all process ΔU1 = ΔU2 = ΔU3
from first law of thermodynamics ΔQ = ΔU + ΔW
Work done ΔW1 > ΔW2 > ΔW3 (Area of P.V. graph)
So ΔQ1 > ΔQ2 > ΔQ3
A gas is taken through the cycle A → B → C → A, as shown in figure. What is the net work done by the gas ? [NEET 2013]
Wnet = Area of triangle ABC =1/2(AC × BC)
During an adiabatic process, the pressure of a gas is found to be proportional to the cube of its temperature. The ratio of 
for the gas is [NEET 2013]
According to question P ∝ T3
But as we know for an adiabatic process the pressure 
A system is taken from state a to state c by two paths adc and abc as shown in the figure. The internal energy at a is Ua = 10 J. Along the path adc the amount of heat absorbed δQ1 = 50 J and the work done δW1 = 20 J whereas along the path abc the heat absorbed δQ2 = 36 J. The amount of work done along the path abc is [NEET Kar. 2013]
From first law of thermodynamics
Again,
Which of the following relations does not give the equation of an adiabatic process, where terms have their usual meaning? [NEET Kar. 2013]
Adiabatic equations of state are:
PVγ = constant
TVγ–1 = constant
P1–γTγ = constant.
Two Carnot engines A and B are operated in series. The engine A receives heat from the source at temperature T1 and rejects the heat to the sink at temperature T. The second engine B receives the heat at temperature T and rejects to its sink at temperature T2. For what value of T the efficiencies of the two engines are equal? [NEET Kar. 2013]
Efficiency of engine A, 
Efficiency of engine B, 
Here, η1 = η2
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