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QUESTION: 1

In a reversible cycle, the entropy of the system

Solution:

QUESTION: 2

Entropy may be expressed as a function of

Solution:

QUESTION: 3

Clausius inequality is stated as

Solution:

If Cycle is reversible

If Cycle is irreversible

If Cycle is impossible

Hence

For cycle is possible

This is known as Clausius inequality.

QUESTION: 4

When a system undergoes a process such that and ΔS > 0, the process is

Solution:

QUESTION: 5

A certain amount of fluid at temperature T_{1} is mixed with an equal amount of the same fluid at temperature T_{2} in an insulated container with total fluid as the system, consider the following statements

I. Energy of the system is conserved

II. Entropy of the system is conserved

III. Entropy of the system increases

IV. Entropy of the system decreases

Q. Which of the above statements is/are true?

Solution:

QUESTION: 6

Which one of the following statements applicable to a perfect gas will also be true for an irreversible process

Solution:

δQ = dU+ PdV... Applicable for a closed system when only PdV work is present. This is true only for a reversible process.

δQ = TdS ...Applicable for a reversible process.

TdS = dU + PdV... Applicable for any process reversible or irreversible, undergone by a closed system, since it is a relation among properties which are independent of the path.

QUESTION: 7

The entropy change for any closed system which undergoes an irreversible adiabatic process

Solution:

QUESTION: 8

Consider two subsystem 1 and 2 containing same fluid and having same mass m; but at Temperature T_{1} and T_{2}(T_{1 }> T_{2}) enclosed in an adiabatic enclosure separate by a partition, if the partition is removed and the fluids are allowed to mix. The entropy change of process is

Solution:

Subsystem 1 having a fluid of mass m_{1} specific heat c_{1} and temperature t_{1} and subsystem 2 consisting of a fluid of mass m_{2}, specific heat c_{2}, and temperature t_{2}, comprise a composite system in an adiabatic enclosure figure. When the partition is removed, the two fluids mix together, and at equilibrium let t_{f }be the final temperature, and t_{2} < t_{f} < t_{1} Since energy interaction is exclusively confined to the two fluids, the system being isolated

∴

Entropy change for the fluid in subsystem1,

This will be negative, since T_{1 }> T_{f}

Entropy change for the fluid in subsystem 2

This will be positive, since T_{2} < T_{f
}

ΔS_{univ} will be positive definite, and the mixing process is irreversible.

Although the mixing process is irreversible, to evaluate the entropy change for the subsystem, the irreversible path was replaced by a reversible path on which the integration was performed.

QUESTION: 9

For the isentropic expansion of an ideal gas from the initial conditions P_{1},T_{1 }to the final conditions P_{2},T_{2}, which one of the following relations is valid?

Solution:

For isentropic expansion, PV^{γ} = constant

from ideal gas law,

Also,

QUESTION: 10

The change in entropy of the system, ΔS_{sys}, undergoing a cyclic irreversible process is

Solution:

Every process proceeds in such direction that total entropy change associated with it will be positive.

QUESTION: 11

In a reversible isothermal process, an ideal gas expands to four times its initial volume. The change in entropy is

Solution:

QUESTION: 12

High pressure steam is expanded adiabatically and reversible through a well insulated turbine which produces some shaft work. If the enthalpy change and entropy change across the turbine are represented by ΔH and ΔS, respectively, for this process:

Solution:

For reversible adiabatic process, ΔS = 0 but ΔH ≠ 0.

QUESTION: 13

A system undergo a state change from 1 to 2, according to second law of thermodynamics for the process to be feasible, the entropy change (S_{2 }- S_{1}) of the system

Solution:

QUESTION: 14

The following four figure have been drawn to represent a fictitious thermodynamic cycle, on P - V and T - S planes

According to first Saw of thermodynamics, equal areas are enclosed by

Solution:

We know that:

For closed cycle change in internal energy is zero First law of thermodynamic for closed system

Hence, equal area are enclosed by figures 1 and 2.

QUESTION: 15

Four process of thermodynamic cycle are shown in figure on P-V diagram in the sequence 1 -2-3-4. The corresponding correct sequence of these process in the T-S plane shown in figure will be

Solution:

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