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# Test: Thermodynamic Processes

## 10 Questions MCQ Test Physics Class 11 | Test: Thermodynamic Processes

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This mock test of Test: Thermodynamic Processes for Class 11 helps you for every Class 11 entrance exam. This contains 10 Multiple Choice Questions for Class 11 Test: Thermodynamic Processes (mcq) to study with solutions a complete question bank. The solved questions answers in this Test: Thermodynamic Processes quiz give you a good mix of easy questions and tough questions. Class 11 students definitely take this Test: Thermodynamic Processes exercise for a better result in the exam. You can find other Test: Thermodynamic Processes extra questions, long questions & short questions for Class 11 on EduRev as well by searching above.
QUESTION: 1

### Which of the following are the extensive variables?

Solution:

Extensive variable →H (enthalpy), E (Internal energy) and Mass. Since these variables depend on the amount of substance or volume or size of the system.

QUESTION: 2

### In an adiabatic process internal energy of gas

Solution:

From the first law of thermodynamics,
we know, dU = dQ - dW ; (work done BY the system is considered +ve)
For an adiabatic process, dQ = 0, and hence, dU = -dW
For an ideal gas expansion, we see that work done
BY the system is +ve (recall the sign convention for work done), i.e., dW > 0.
Therefore, dU is less than 0, and thus, the internal energy decreases.

QUESTION: 3

### Find the final temperature of one mole of an ideal gas at an initial temperature to t K.The gas does 9 R joules of work adiabatically. The ratio of specific heats of this gas at constant pressure and at constant volume is 4/3.

Solution:

TInitial  = t K
Work, W = 9R
Ratio of specific heats, γ = C/ Cv = 4/3
In an adiabatic process, we have
W = R(TFinal – Tinitial) / (1-γ)
9R = R (TFinal – t) / (1 – 4/3)
TFinal – t = 9 (-1/3) = -3
TFinal  = (t-3) K

QUESTION: 4

In an adiabatic process gas is reduced to quarter of its volume. What would happen to its pressure? Given ratio of specific heats γ= 2

Solution:

As given the ratio of specific heats = 2, we get PV2 is constant
Hence when the volume is decreased to its quarter, for PV2  to be constant P must increase by 16 times.

QUESTION: 5

In an open system, for maximum work, the process must be entirely

Solution:

A reversible process gives the maximum work.

QUESTION: 6

Molar specific heat of a substance denoted by symbol C does not depend upon

Solution:

The symbol c stands for specific heat and depends on the material and phase. The specific heat is the amount of heat necessary to change the temperature of 1.00 kg of mass by 1.00degreeC. The specific heat c is a property of the substance; its SI unit is J/(kg⋅K) or J/(kg⋅C).

QUESTION: 7

Two gases X and Y kept in separate cylinders with same initial temperature and pressure are compressed to one third of their volume through isothermal and adiabatic process respectively. Which gas would have more pressure?

Solution:
QUESTION: 8

Heat is supplied to the gas, but its internal energy does not increase. What is the process involved?

Solution:

From the first law of thermodynamics  dq=du+dw,so clearly for the isothermal expansion or compression of a real gas where u=f(T) from the first law du=0 which means that the entire heat supplied is  converted into work but from the second law of thermodynamics we find that in no,process can the entire heat supplied can  be converted into work hence in reality some fraction of heat supplied is always used to increase the internal energy of the system.

QUESTION: 9

Minimum work is said to be done when a gas expands

Solution:

An isobaric expansion of a gas requires heat transfer to keep the pressure constant. An isochoric process is one in which the volume is held constant, meaning that the work done by the system will be zero.

QUESTION: 10

What is not true for a cyclic process?

Solution:

As work is a path function rather than a state function, we can easily say that work can often be graphically represented as the area under the PV graph. And as cyclic processes are represented as closed shapes on PV graph it is obvious that they have non zero area and thus work done is non zero.