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QUESTION: 1

In thermodynamic processes which of the following statements is not true? [2009]

Solution:

In an isochoric process volume remains constant whereas pressure remains constant in isobaric process.

QUESTION: 2

The internal energy change in a system that has absorbed 2 kcals of heat and done 500 J of work is: [2009]

Solution:

According to first law of thermodynamics Q = ΔU + W

ΔU = Q – W

= 2 × 4.2 × 1000 – 500 = 8400 –500

= 7900 J

QUESTION: 3

If ΔU and ΔW repr esent the in crease in internal energy and work done by the system respectively in a thermodynamical process, which of the following is true? [2010]

Solution:

By first law of thermodynamics, ΔQ = ΔU + ΔW

In adiabatic process, ΔQ = 0

∴ ΔU = -ΔW

In isothermal process, ΔU = 0

∴ ΔQ -ΔW

QUESTION: 4

During an isothermal expansion, a confined ideal gas does –150 J of work against its surroundings. This implies that [2011]

Solution:

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.

QUESTION: 5

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]

Solution:

Change in entropy is given by

QUESTION: 6

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]

Solution:

T_{1} = 273 + 27 = 300K

T_{2} = 273 + 927 = 1200K

For adiabatic process,

= P_{1} (2^{7}) = 2 × 128 = 256 atm

QUESTION: 7

A thermodynamic system is taken through the cycle ABCD as shown in figure. Heat rejected by the gas during the cycle is : [2012]

Solution:

∵ 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

QUESTION: 8

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]

Solution:

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)

QUESTION: 9

If Q_{1}, Q_{2}, Q_{3} indicate the heat a absorbed by the gas along the three processes and ΔU_{1}, ΔU_{2}, ΔU_{3} indicate the change in internal energy along the three processes respectively, then

Solution:

Initial and final condition is same for all process ΔU_{1 }= ΔU_{2} = ΔU_{3}

from first law of thermodynamics ΔQ = ΔU + ΔW

Work done ΔW_{1} > ΔW_{2} > ΔW_{3} (Area of P.V. graph)

So ΔQ_{1} > ΔQ_{2} > ΔQ_{3}

QUESTION: 10

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]

Solution:

Wnet = Area of triangle ABC AC × BC

QUESTION: 11

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]

Solution:

According to question P ∝ T^{3}

But as we know for an adiabatic process the pressure

QUESTION: 12

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 U_{a} = 10 J. Along the path adc the amount of heat absorbed δQ_{1} = 50 J and the work done δW_{1} = 20 J whereas along the path abc the heat absorbed δQ_{2 }= 36 J. The amount of work done along the path abc is [NEET Kar. 2013]

Solution:

From first law of thermodynamics

Again,

QUESTION: 13

Which of the following relations does not give the equation of an adiabatic process, where terms have their usual meaning?

Solution:

Adiabatic equations of state are PV^{γ} = constant

TV^{γ–1} = constant

P^{1–γ}T^{γ} = constant.

QUESTION: 14

Two Carnot engines A and B are operated in series. The engine A receives heat from the source at temperature T_{1} 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 T_{2}. For what value of T the efficiencies of the two engines are equal? [NEET Kar. 2013]

Solution:

Efficiency of engine A,

Efficiency of engine B,

Here, η_{1} = η_{2}

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