Q#9
Can work be done by a system without changing its volume?Answer:
If the system is not changing its volume the process is called isochoric. In this process, work can not be done because ΔW = pΔV = p x 0 = 0.
Q#10
An ideal gas is pumped into a rigid container having diathermic walls so that the temperature remains constant. In a certain time interval, the pressure in the container is doubled. Is the internal energy of the contents of the container also doubled in the interval?
Answer:
n = pV/RT, here, V, R, and T are constant, so when pressure is 2p, the number of moles = 2n.
Since the mass or number of moles of the gas is doubled, the internal energy associated with the original mass is also doubled in the contents of the cylinder.
Q#11
When a tire bursts, the air coming out is cooler than the surrounding air. Explain.
Answer:
The tire burst is a sudden process in which heat transfer between the system and the surrounding can not take place. So it is an adiabatic process. The pressure of the air inside the tire is more than the surrounding. Due to the tire burst, the volume of the air increases and some work is done by the air for which it uses its internal energy. So the internal energy of the system decreases. It results in a decrease in the temperature of the expanded air and feels cooler than the surrounding.
Q#12
When we heat an object, it expands. Is the work done by the object in this process? Is heat given to the object equal to the increase in its internal energy?
Answer:
ΔW = pΔV
Not only gases, but it is also true for the expansion of solids and liquids.
In the given problem, p is constant but due to expansion ΔV has some positive value. Hence some work is done by the object.
Now the change in the internal energy,
ΔU = ΔQ - ΔW
Since ΔW is not zero, ΔU < ΔQ.
So, the heat given to the object is not equal to the increase in the internal energy.
Q#13
When we stir a liquid vigorously, it becomes warm. Is it a reversible process?
Answer:
When a liquid is vigorously stirred it becomes warm because the work is done against the viscosity. But the process is dissipative. A dissipative process can not be reversible. Another point is that if mechanical work is done on the system to increase the internal energy the reverse process is not automatic but we need a heat engine to do it that also only a part of the heat energy can be used to do the mechanical work. So, it is not a reversible process.
Q#14
What should be the condition for the efficiency of a Carnot engine to be equal to 1?
Answer:
The efficiency of a Carnot engine, η = 1 - Q₂/Q₁, where Q₁ is the heat applied to the engine and Q₂ is the heat given to the sink. It is clear that η can be equal to 1 only when Q₂ is zero but it is not possible according to the second law of thermodynamics which is also called Kelvin-Planck statement.
Q#15
When an object cools down, heat is withdrawn from it. Does the entropy of the object decrease in the process? If yes, is it a violation of the second law of thermodynamics stated in terms of an increase in entropy?
Answer:
Yes, when the heat is withdrawn from an object its entropy decreases.
The second law of thermodynamics may be stated as "It is not possible to have a process in which the entropy of an isolated system is decreased". Here the cooling object is not an isolated system. The heat given out by the object is added to the surroundings and its entropy increases. The universe along with the object is taken as an isolated system. And its entropy is not decreasing. Hence the second law of thermodynamics is not violated.
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