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# Test: Thermodynamic Processes

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## 15 Questions MCQ Test Topic-wise MCQ Tests for NEET | Test: Thermodynamic Processes

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Test: Thermodynamic Processes - Question 1

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

Detailed Solution for Test: Thermodynamic Processes - Question 1

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.

Test: Thermodynamic Processes - Question 2

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

Detailed Solution for Test: Thermodynamic Processes - Question 2

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.

Test: Thermodynamic Processes - Question 3

### What is not true for a cyclic process?

Detailed Solution for Test: Thermodynamic Processes - Question 3

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.

Test: Thermodynamic Processes - Question 4

In an adiabatic process internal energy of gas

Detailed Solution for Test: Thermodynamic Processes - Question 4

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.

Test: Thermodynamic Processes - Question 5

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.

Detailed Solution for Test: Thermodynamic Processes - Question 5

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

Test: Thermodynamic Processes - Question 6

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

Detailed Solution for Test: Thermodynamic Processes - Question 6

Explanation : The adiabatic condition is given by the relation between pressure volume and temperature volume as:

(PV)γ = constant

where, γ = Cp/Cv is ratio of the specific heats

These relations suggest that an decrease in volume is associated with increase in temperature

ATQ

P1(V1)γ = (P2V2

=> P1(1)2 = P2(4)2

P1/P2 = 16

Test: Thermodynamic Processes - Question 7

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

Detailed Solution for Test: Thermodynamic Processes - Question 7

A reversible process gives the maximum work.

Test: Thermodynamic Processes - Question 8

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

Detailed Solution for Test: Thermodynamic Processes - Question 8

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).

Test: Thermodynamic Processes - Question 9

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?

Test: Thermodynamic Processes - Question 10

In which of the following processes is heat transfer equal to zero?

Detailed Solution for Test: Thermodynamic Processes - Question 10

Entropy is defined as dQ/T. In an isentropic process dQ/T = 0 which means dQ = 0. Diathermic is a process in which heat flow is easily possible. Isochoric process is one in which volume is constant. Isothermal process is one in which temperature stays constant.

Test: Thermodynamic Processes - Question 11

Minimum work is said to be done when a gas expands

Detailed Solution for Test: Thermodynamic Processes - Question 11

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.

Test: Thermodynamic Processes - Question 12

Which of the following variables is zero for a cyclic process?

Detailed Solution for Test: Thermodynamic Processes - Question 12

In a cyclic process, the starting position is the same as the ending position. So, the change in all state variables is zero. So the net change in internal energy is zero. ΔU = ΔQ – ΔW = 0.

Test: Thermodynamic Processes - Question 13

Isothermal process can be represented by which law?

Detailed Solution for Test: Thermodynamic Processes - Question 13

In an isothermal process, PV=const. This is the same as Boyle’s law. Charle’s law is given by: V/T=const. Gay-Lussac’s law is given by: P/T=const and 2nd law of thermodynamics states that in every process total entropy of the universe must increase.

Test: Thermodynamic Processes - Question 14

Calculate the work done by the gas in an isothermal process from A to B. PA = 1Pa, VA = 3m3, PB = 3Pa.

Detailed Solution for Test: Thermodynamic Processes - Question 14

Since the process is isothermal the product PV will be constant.
PAVA = PBVB.
∴ VB = 1*3/3 = 1m3.
Work done in an isothermal process is given by:
nRT*ln(VB/VA)
= PAVAln(VB/VA)
= 3 ln(1/3)
= - 3.3 J.

Test: Thermodynamic Processes - Question 15

The given graph corresponds to which equation?

Detailed Solution for Test: Thermodynamic Processes - Question 15

In the given graph temperature remains constant with variation in volume. So the process is isothermal and PV = constant.

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