Questions OBJECTIVE - I and Answer Laws of Thermodynamics HC Verma Part II
H C Verma solutions, Laws of Thermodynamics, OBJECTIVE-I, Chapter-26, Concepts of Physics, Part-II
Laws of Thermodynamics
OBJECTIVE-I
Q#1
The first law of thermodynamics is a statement of
(a) Conservation of heat
(b) Conservation of work
(c) Conservation of momentum
(d) Conservation of energy.
Answer: (d)
EXPLANATION: The statement of the first law of thermodynamics is,
ΔQ = ΔU + ΔW.
i.e. if ΔQ amount of heat is given to a gas in a certain process and an amount of ΔW work is done by it then it's internal energy must increase by an amount of ΔQ -ΔW. So it is just a restatement of the law of conservation of energy.
Q#2
If heat is supplied to an ideal gas in an isothermal process,
(a) the internal energy of the gas will increase
(b) the gas will do positive work
(c) the gas will do negative work
(d) the said process is not possible.
Answer: (b)
EXPLANATION: Since the process is isothermal there is no change in the temperature. Also, the internal energy is a function of temperature, there is no change in internal energy. Means ΔU =0. From the first law of thermodynamics, ΔQ =ΔU+ΔW
→ΔQ = ΔW.
So, the gas will do positive work.
Q#3
Figure (26-Q1) shows two processes A and B on a system. Let ΔQ₁ and ΔQ₂ be the heat given to the system in processes A and B respectively. Then
(a) ΔQ₁ > ΔQ₂
(b) ΔQ₁= ΔQ₂
(c) ΔQ₁ < ΔQ₂
(d) ΔQ₁ ≤ ΔQ₂.
Explanation: Since the initial and final pressure and volume are the same for both the processes the initial and final temperatures of the system will also be the same. Thus the change in the internal energy (ΔU) of the system will also be the same for both the processes.
From the first law of thermodynamics, ΔQ =ΔU+ΔW. Since U is the same for both the processes hence ΔQ will be greater for greater ΔW. From the p-V graph, the area under the curve is equal to the work done. Since the area under the process-A is more than the area under process B, ΔW₁ > ΔW₂. Thus, ΔQ₁ > ΔQ₂.
Q#4
Refer to the figure (26-Q1). Let ΔU₁ and ΔU₂ be the changes in internal energy of the system in the processes A and B. Then
(a) ΔU₁ > ΔU₂
(b) ΔU₁ = ΔU₂
(c) ΔU₁ < ΔU₂
(d) ΔU₁ ≠ ΔU₂.
Answer: (b)
Explanation: Since the initial and final pressure and volume are the same for both the processes the initial and final temperatures of the system will also be the same. Thus the change in the internal energy (ΔU) of the system will also be the same for both the processes.
Q#5
Consider the process on a system shown in figure (26-Q2). During the process, the work done by the system
(a) continuously increases
(b) continuously decreases
(c) first increases then decreases
(d) first decreases then increases.
Explanation: The work done by the system is equal to the area below the curve in the p-V graph. In the figure (36-Q2), during the process area under the curve continuously increases from the starting point the work done by the system continuously increases.
Q#6
Consider the following two statements.
(A) If heat is added to a system, its temperature must increase.
(B) If positive work is done by a system in a thermodynamic process, its volume must increase.
(a) Both A and B are correct.
(b) A is correct but B is wrong.
(c) B is correct but A is wrong.
(d) Both A and B are wrong.
Answer: (c)
Explanation: If heat is added to the system its temperature depends upon the work done by the system. If the work done by the system ΔW < ΔQ then the temperature of the system will increase but if ΔW = ΔQ then it will not increase. Hence the statement A is not true.
The positive work done by the system is the area under the p-V curve. i.e. it is equal to ∫p*dV. If Dv = 0, then work done by the system is zero. For a positive work, dV must have positive value i.e. the volume must increase. Statement B is true.
Q#7
An ideal gas goes from the state i to the state f as shown in figure (26-Q3). The work done by the gas during the process
(a) is positive
(b) is negative
(c) is zero
(d) cannot be obtained from this information
Answer: (c)
Explanation: The p-T graph is a straight line, so the pressure is directly proportional to the temperature.
Since for a closed system, p =nRT/V,
Here p ∝ T, so nR/V = constant. Here n and R are constants, so V is also constant. Since V does not change the work done is zero because work done =∫p.dV.
Q#8
Consider two processes on a system as shown in figure (26-Q4).
(a) ΔW₁ > ΔW₂
(b) ΔW₁ = ΔW₂
(c) ΔW₁ < ΔW₂
(d) Nothing can be said about the relation between ΔW₁ and ΔW₂.
Answer: (c)
Explanation: Since the initial and final volumes in both the processes are the same, the change in volume ΔV is the same. The work done by the system in the process A, ΔW₁ =p₁*ΔV and in process B, ΔW₂ = p₂*ΔV. From the graph, p₁ < p₂, so ΔW₁ < ΔW₂.
Q#9
A gas is contained in a metallic cylinder fitted with a piston. The piston is suddenly moved in to compress the gas and is maintained at this position. As time passes the pressure of the gas in the cylinder
(a) increases
(b) decreases
(c) remains constant
(d) increases or decreases depending on the nature of the gas.
Answer: (b)
Explanation: Since the gas is suddenly compressed, the process is adiabatic. Both the pressure and temperature increase. Since the cylinder is metallic the heat of the gas gets out slowly through the walls. The volume is now constant so no work done. Hence due to the loss of heat energy the temperature slowly decreases. For a closed system,
pV/T =constant. If V is also constant then, p ∝ T. Since T decreases, p also decreases.
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