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ISRO 2007 Mechanical Questions, Answers and Explanation

ISRO 2007 Mechanical Questions, Answers and Explanation

1. Only rocket engines can be propelled to space because
a) They can generate very high thrust
b) They have high propulsion efficiency
c) These engines can work on several fuels
d) They are not air-breathing engines

Answer
d) They are not air-breathing engines

Explanation [Engineering Mechanics]
Even though all the choices are true, for travel in space the engines must use their own oxidizer. So they should not be dependent on air

2. The universal gas constant of a gas is the product of molecular weight of the gas and 
a) Gas constant
b) Specific heat at constant pressure
c) Specific heat at constant volume
d) None of the above

Answer
a) Gas constant

Explanation [Thermal Engineering]
The value of universal gas constant is 8.314 J mol-1K-1. This when divided by the molecular weight of gas we get specific gas constant
Molecular mass of air = 28.97 g/mol
Rair = 8.314x1000/28.97 J kg-1K-1 = 287  J kg-1K-1.

3. Van der Waal's equation of state of a gas is
a) pV = nRT
b) (p+a/V2,/sup>)(v+b) = RT
c) (p+a/V2,/sup>)(v-b) = RT
d) (p-a/V2,/sup>)(v-b) = RT

Answer
c) (p+a/V2,/sup>)(v-b) = RT

Explanation [Thermal Engineering]
The constant 'a' provides correction for inter molecular forces and b for finite molecular size.  The effect of the finite volume of the particles is to decrease the available void space in which the particles are free to move. We must replace V by V − b, where b is called the excluded volume or "co-volume". 
The pressure of a gas is due to the hits of the molecules on the walls of the containing vessel. The attractive force between the molecules comes into play when the molecules are brought close together by compressing the gas. A molecule in the body of the gas is attracted in all the directions when forces acting in opposite directions cancel out, but a molecule, the boundary of the gas is subjected to an inward pull due to unbalanced molecular attraction. In this way some of the energy of the molecule about to strike the wall of the vessel is used up in overcoming this inward pull. Therefore, it will not strike the opposite wall with the same force. The observed pressure consequently will be less than the ideal pressure. Therefore the ideal pressure, will be equal to observed pressure P plus a pressure correction, depending upon the attractive forces.


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