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Question-1

What do you mean by a closed system?

Solution:
If a system exchanges neither energy nor matter with its surroundings then it is called as a closed system. A system is said to be an open system if it can exchange both energy and matter with its surroundings.

Question-2

What is the mathematical form of first law of thermodynamics?

Solution:
The first law of thermodynamics is the law of conservation of energy. It states that if a system absorbs heat 'dQ' and as a result, the internal energy of the system changes by 'dU' and the system does a work 'dW', then

                  dQ = dU + dW

But             dW = PdV

Therefore,   dQ = dU dV + PdV

which is the mathematical form of first law of thermodynamics.

Question-3

The efficiency of heat engine cannot be 100%. Explain why?

Solution:
The efficiency of heat engine is given by

η= 1 - h = 1 -
Q2 = Heat rejected to the sink
Q1 = Heat absorbed from the source


η= 1 or 100% efficiency if and only if Q2 = 0

This cannot happen because if Q2 = 0, then the temperature of the working substance will go on increasing. A stage will come when the temperature of the working substance becomes equal to the temperature of the source.

Question-4

What are the limitations of the first law of thermodynamics?

Solution:
The following are the limitations of the first law of thermodynamics:-

(i) The first law does not tell us about the direction of flow of heat.

(ii) It fails to explain why heat cannot be spontaneously converted into work.

Question-5

What are the basic requirements of cooking utensils with respect to specific heat, thermal conductivity and coefficient of expansion?

Solution:
A cooking utensil should be of suitable material, which possesses small value of specific heat (c), large value of thermal conductivity (k) and low coefficient of expansion.

Question-6

Oxygen cannot be liquefied at 00C, however large be the applied pressure. However CO2 at 00C can be liquefied by applying sufficient pressure. Why?

Solution:
The critical temperature of oxygen is 1180C. At 00C, oxygen is much above its critical temperature. Therefore, it cannot be liquefied, however large be the applied pressure. On the other hand, the critical temperature of CO2 is 310C. CO2 at 00C is below its critical temperature. Hence, it can be liquefied by the application of sufficient pressure.

Question-7

If you were asked to increase the efficiency of a Carnot engine by increasing the temperature of the source or by decreasing the temperature of the sink by 10K,which would you prefer and why?

Solution:

To increase the efficiency of the Carnot engine, decreasing the temperature of the sink would be preferable. This is because; decreasing the temperature of the sink by 10K brings would bring about greater increase in the efficiency of the heat engine than what is achieved by increasing the temperature of the source by 10K.

Question-8

A petrol engine consumes 25kg of petrol per hour. The calorific value of petrol is 11.4 × 106 cal Kg-1. The power of the engine is 99.75 KW. Calculate the efficiency of the engine.

Solution:
m = 25kg/hour
Calorific value = 11.4
× 106 cal kg-1
Output power = 94.75 KW = 99.75 × 103 W
Efficiency,
η= ?
Heat produced/hour = 11.4 ×106 × 25 cal
                            = 11.4 × 106 × 25 × 4.2 J



Input power = W
                   

Question-9

Calculate the difference between two principal specific heats of 1g of helium gas at N. T. P. molecular mass of helium = 4 and J = 4.186Jcal-1 and R = 8.31 J mol-1K-1.

Solution:
Cp - Cv = 7
Molecular mass, M = 4
R = 8.31J mol-1K-1

Question-10

Which molecules, ice at 00C or water at 00C , have greater potential energy and why?

Solution:
The potential energy of water molecules at 00C is more, because heat spent in melting is used up in increasing the potential energy.

Question-11

What do you mean by open system?

Solution:
A system is said to be an open system if it can exchange both energy and matter with its surroundings.

Question-12

Which molecules, ice at 00 C or water at 00 C has greater potential energy and why?

Solution:
The potential energy of water molecules at 00 C is more, because heat spent in melting is used up in increasing the P.E.

Question-13

Does the internal energy of an ideal gas change in an isothermal process and adiabatic process?

Solution:
In an isothermal process, T = constant
ΔT = 0
dU = 0
In an adiabatic process, Q = constant
dQ = 0, therefore, dU = - dW
0.

Question-14

What happens to the phase of water at its critical point?

Solution:
At the critical point, water and water vapours are equally dense.

Question-15

What is the efficiency of a Carnot engine operating between boiling and freezing point of water?

Solution:

Question-16

Why is a gas cooled when expanded?

Solution:
This is due to decrease in internal energy.

Question-17

At what temperature and pressure can the solid, liquid and vapour phases of CO2 co-exist in equilibrium?

Solution:
At the triple point, temperature = -56.60 C and pressure = 5.11 atm.

Question-18

Why does the pressure increase when the temperature of a gas increases?

Solution:
When temperature of gas increases, the K.E. of the molecules increases. Therefore, the molecules strike against the walls of the container with large momentum. Due to this the pressure increases.

Question-19

The efficiency of heat engine cannot be 100%. Explain why?

Solution:
The efficiency of heat engine is given by η = 1 – Q2 / Q1

Q1 is the heat rejected to the sink
Q2 is the heat absorbed from the source
The coefficient of friction
η will be 100% only when Q2 is zero
Q2 = 0 is not possible because the temperature of the working substance will increase and a stage occurs that the temperature of the working substance becomes equal to the temperature of the source. In this situation, there is no transfer of heat from the source to the working substances and there will be no output.

Question-20

What happens when CO2 at 4 atm pressure is cooled from room temperature at constant pressure?

Solution:
It condenses to a solid state directly without passing through the liquid phase.

Question-21

Two rods of the same length, and with a rectangular cross-section (area of the cross-section being C1 and C2) are joined parallel to each other. Their thermal conductivities are K1 and k2. Show that the combination is equivalent to a material of conductivity.

K=.


Solution:
For the upper rod, Q1 =
For the lower rod, Q2 =
Total amount of heat flowing per unit time = Q1 + Q2
Total area of the cross section = A1 + A2
If K is equivalent conducting

We have, Q1 + Q2 =
                           =
or, K(A1 + A2) = K1A1 + K2A2      or




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