- #1

- 170

- 14

Hello again, all! I have another basic thermodynamics question :P This one came from a 2008 admission test. Seems simple enough, but as usual, my answer doesnt match =P

A monoatomic ideal gas performs, reversibly, the cycle shown in the diagram attached in this post. The values of [tex]P_0[/tex] and [tex]V_0[/tex] are, respectively, [tex]1\times 10^5 Pa[/tex] and [tex]100cm^3[/tex]. The area of the interior of the cycle is 15J. What is the efficiency of this cycle?

[tex]e=\frac{W}{Q_h}[/tex]

e is the efficiency, W is work done by the system and [tex]Q_h[/tex] is the energy absorbed by the system (heat).

This is probably the only equation relevant here..

The problem seemed straigh forward: The area enclosed by the cycle is the work done. The area under the curve [tex]ab[/tex] is the [tex]Q_h[/tex], right? If so, calculating this is trivial and results in an efficiency of, approximately, [tex]0,43[/tex].

However, this is not right. The correct answer should be [tex]0,136[/tex].

## Homework Statement

A monoatomic ideal gas performs, reversibly, the cycle shown in the diagram attached in this post. The values of [tex]P_0[/tex] and [tex]V_0[/tex] are, respectively, [tex]1\times 10^5 Pa[/tex] and [tex]100cm^3[/tex]. The area of the interior of the cycle is 15J. What is the efficiency of this cycle?

## Homework Equations

[tex]e=\frac{W}{Q_h}[/tex]

e is the efficiency, W is work done by the system and [tex]Q_h[/tex] is the energy absorbed by the system (heat).

This is probably the only equation relevant here..

## The Attempt at a Solution

The problem seemed straigh forward: The area enclosed by the cycle is the work done. The area under the curve [tex]ab[/tex] is the [tex]Q_h[/tex], right? If so, calculating this is trivial and results in an efficiency of, approximately, [tex]0,43[/tex].

However, this is not right. The correct answer should be [tex]0,136[/tex].