Chapter 2, Problem 2.6E

 Consider exercise 2.5. Would the work be more or less if it were performed against different external pressures found (a) at the top of Mount Everest, (b) at the bottom of Death Valley, (c) in space?

Interpretation Introduction

(a)

Interpretation:

Predict whether the work will be more or less when it is performed at the top of the Mount Everest.

Concept introduction:

The work is performed on an object when an object moves a certain distance s due to the force F. Mathematically, it is indicated by the dot product of the force vector F and the distance vector s. The mathematical equation is given below,

$\mathrm{W}\mathrm{o}\mathrm{r}\mathrm{k}\mathrm{ }=\mathrm{ }\mathrm{F}.\mathrm{s}\mathrm{ }=\mathrm{ }\mathrm{ }\left|\mathrm{F}\right|\left|\mathrm{s}\right|\mathrm{ }\mathrm{c}\mathrm{o}\mathrm{s}\mathrm{ }Ѳ$

Where Ѳ is the angle between the vectors F and s. The unit of work is joules. Work is a way to transfer the energy. The energy is defined as the ability to do work and so energy and work are described using the same unit in joules.

Answer to Problem 2.6E

The work would be less because the external pressure is less.

Explanation of Solution

Gases do expansion or compression work by following the equation.

$\mathrm{W}\mathrm{o}\mathrm{r}\mathrm{k}\mathrm{ }\mathrm{d}\mathrm{o}\mathrm{n}\mathrm{e}\mathrm{ }\mathrm{o}\mathrm{n}\mathrm{ }\mathrm{t}\mathrm{h}\mathrm{e}\mathrm{ }\mathrm{s}\mathrm{y}\mathrm{s}\mathrm{t}\mathrm{e}\mathrm{m}\mathrm{ }=\mathrm{ }-\mathrm{P}∆\mathrm{V}$

The given statement is like a person lifting a book or any other material from ground state to higher state does work against the gravity.

The gases can do the work through expansion or compression against the constant external pressure. Thus when the gas does the work in a piston and move it against the external pressure. Then the volume increase and so the work is negative as the external pressure is less.

Conclusion

Thus the work would be less because the external pressure is less.

Interpretation Introduction

(b)

Interpretation:

Predict whether the work will be more or less when it is performed at the bottom of the Death valley.

Concept introduction:

The work is performed on an object when an object moves a certain distance s due to the force F. Mathematically, it is indicated by the dot product of the force vector F and the distance vector s. The mathematical equation is given below,

$\mathrm{W}\mathrm{o}\mathrm{r}\mathrm{k}\mathrm{ }=\mathrm{ }\mathrm{F}.\mathrm{s}\mathrm{ }=\mathrm{ }\mathrm{ }\left|\mathrm{F}\right|\left|\mathrm{s}\right|\mathrm{ }\mathrm{c}\mathrm{o}\mathrm{s}\mathrm{ }\mathrm{Ѳ}$

Where Ѳ is the angle between the vectors F and s. The unit of work is joules. Work is a way to transfer the energy. The energy is defined as the ability to do work and so energy and work are described using the same unit in joules.

Answer to Problem 2.6E

$\mathrm{T}\mathrm{y}\mathrm{p}\mathrm{e}\mathrm{ }\mathrm{e}\mathrm{q}\mathrm{u}\mathrm{a}\mathrm{t}\mathrm{i}\mathrm{o}\mathrm{n}\mathrm{ }\mathrm{h}\mathrm{e}\mathrm{r}\mathrm{e}.$

The work would be more because the external pressure is more.

Explanation of Solution

Gases do expansion or compression work by following the equation.

$\mathrm{W}\mathrm{o}\mathrm{r}\mathrm{k}\mathrm{ }\mathrm{d}\mathrm{o}\mathrm{n}\mathrm{e}\mathrm{ }\mathrm{o}\mathrm{n}\mathrm{ }\mathrm{t}\mathrm{h}\mathrm{e}\mathrm{ }\mathrm{s}\mathrm{y}\mathrm{s}\mathrm{t}\mathrm{e}\mathrm{m}\mathrm{ }=\mathrm{ }-\mathrm{P}∆\mathrm{V}$

The given statement is like a person lifting a book or any other material from ground state to same state does work with the gravity.

The gases can do the work through expansion or compression against the constant external pressure. Thus when the gas does the work in a piston and move it with the external pressure. Then the volume decrease and so the work is positive as the external pressure is more.

Conclusion

Thus the work would be more because the external pressure is more.

Interpretation Introduction

(c)

Interpretation:

Predict whether the work will be more or less when it is performed at space.

Concept introduction:

The work is performed on an object when an object moves a certain distance s due to the force F. Mathematically, it is indicated by the dot product of the force vector F and the distance vector s. The mathematical equation is given below,

$\mathrm{W}\mathrm{o}\mathrm{r}\mathrm{k}\mathrm{ }=\mathrm{ }\mathrm{F}.\mathrm{s}\mathrm{ }=\mathrm{ }\mathrm{ }\left|\mathrm{F}\right|\left|\mathrm{s}\right|\mathrm{ }\mathrm{c}\mathrm{o}\mathrm{s}\mathrm{ }\mathrm{Ѳ}$

Where Ѳ is the angle between the vectors F and s. The unit of work is joules. Work is a way to transfer the energy. The energy is defined as the ability to do work and so energy and work are described using the same unit in joules.

Answer to Problem 2.6E

No work would be performed because the external pressure is zero.

Explanation of Solution

Gases do expansion or compression work by following the equation.

$\mathrm{W}\mathrm{o}\mathrm{r}\mathrm{k}\mathrm{ }\mathrm{d}\mathrm{o}\mathrm{n}\mathrm{e}\mathrm{ }\mathrm{o}\mathrm{n}\mathrm{ }\mathrm{t}\mathrm{h}\mathrm{e}\mathrm{ }\mathrm{s}\mathrm{y}\mathrm{s}\mathrm{t}\mathrm{e}\mathrm{m}\mathrm{ }=\mathrm{ }-\mathrm{P}∆\mathrm{V}$

Considering the expansion of a gas into a larger volume, which is an initially a vacuum. In this, the gas is expanding against a P_ext of 0, where the work done by the gas is equals to zero. Such a process is called a free expansion. Hence, work = 0 for free expansion.

Conclusion

Thus no work would be performed because the external pressure is zero.