# Pressure-Volume Diagrams

## Practice

### practice problem 1

- Sketch the PV graph of this cycle.
- Determine the temperature at state A, B, C, and D.
- Calculate
*W*,*Q*, and Δ*U*on the path A→B, B→C, C→D, D→A and for one complete cycle. (Include the algebraic sign with each value.) - Does this cycle behave more like an engine or a refrigerator?

state | A | B | C | D | ||
---|---|---|---|---|---|---|

P |
(Pa) | 100,000 | 100,000 | 200,000 | 200,000 | |

V |
(m^{3}) |
0.020 | 0.060 | 0.060 | 0.020 | |

T |
(K) | |||||

path | A→B | B→C | C→D | D→A | ABCDA | |

description | isobaric | isochoric | isobaric | isochoric | closed cycle | |

ΔU |
(J) | |||||

Q |
(J) | |||||

W |
(J) |

#### solution

Plot the four given points. Connect them with straight lines. Isobars are horizontal lines. Isochors are vertical lines. Add arrows to show the direction of each path. The overall cycle runs counterclockwise.

Compute the temperature at each point using the ideal gas law.

*T*=*PV*⇐ *PV*=*nRT**nR*Do it four times. Note that the temperature changes between points are all simple multiples (2×, 3×) or simple fractions (½, ⅓). That's because the pressures and volumes always changed by simple amounts. The numbers were rigged to make it easy to check your work.

*T*=_{A}(100,000 Pa)(0.020 m ^{3})= 240 K (1 mol)(8.31 J/mol K) *T*=_{B}(100,000 Pa)(0.060 m ^{3})= 720 K (1 mol)(8.31 J/mol K) *T*=_{C}(200,000 Pa)(0.060 m ^{3})= 1440 K (1 mol)(8.31 J/mol K) *T*=_{D}(200,000 Pa)(0.020 m ^{3})= 480 K (1 mol)(8.31 J/mol K) Compute the change in internal energy on each segment using the function of state.

∆

*U*=^{3}_{2}*nR*∆*T*Do this four times, and watch the signs. They matter for the computations that follow.

∆ *U*_{A→B}=^{3}_{2}(1 mol)(8.31 J/mol K)(720 K − 240 K)∆ *U*_{B→C}=^{3}_{2}(1 mol)(8.31 J/mol K)(1440 K − 720 K)∆ *U*_{C→D}=^{3}_{2}(1 mol)(8.31 J/mol K)(480 K − 1440 K)∆ *U*_{D→A}=^{3}_{2}(1 mol)(8.31 J/mol K)(240 K − 480 K)Let's start filling in the table. Include a sign with all numbers associated with a path. The pressure volume, and temperature numbers are always positive, so adding a sign to them would be redundant. I like to use red for negative values in a table, but that's just a personal preference.

A four step cycle (isobaric & isochoric) state A B C D *P*(Pa) 100,000 100,000 200,000 200,000 *V*(m ^{3})0.020 0.060 0.060 0.020 *T*(K) 240 720 1440 480 path A→B B→C C→D D→A ABCDA description isobaric isochoric isobaric isochoric closed cycle Δ *U*(J) +6,000 +9,000 −12,000 −3,000 *Q*(J) *W*(J) Compute the work done on each segment next. That's just an area with a sign switch. The areas under the isochors are both zero. The areas under the isobars are both rectangles. When you find an area under a segment running in the positive direction it's positive. When you go the other way, it's negative. If you prefer to do calculations systematically, compute the rectangular areas using base times height, where "base" is final minus initial volume (order of subtraction matters) and "height" is volume.

*W*= −*area*= −*bh*= −(*V*_{2}−*V*_{1})*P*Do this twice.

*W*_{A→B}=−(0.060 m ^{3}− 0.020 m^{3})100,000 Pa*W*_{C→D}=−(0.020 m ^{3}− 0.060 m^{3})200,000 PaAdd your results to the table.

A four step cycle (isobaric & isochoric) state A B C D *P*(Pa) 100,000 100,000 200,000 200,000 *V*(m ^{3})0.020 0.060 0.060 0.020 *T*(K) 240 720 1440 480 path A→B B→C C→D D→A ABCDA description isobaric isochoric isobaric isochoric closed cycle Δ *U*(J) +6,000 +9,000 −12,000 −3,000 *Q*(J) *W*(J) −4,000 0 +8,000 0 The remaining empty cells are filled in using the first law of thermodynamics…

∆

*U*=*Q*+*W*or, in the case of the whole cycle ABCD, by just doing a sum across a row.

A four step cycle (isobaric & isochoric) state A B C D *P*(Pa) 100,000 100,000 200,000 200,000 *V*(m ^{3})0.020 0.060 0.060 0.020 *T*(K) 240 720 1440 480 path A→B B→C C→D D→A ABCDA description isobaric isochoric isobaric isochoric closed cycle Δ *U*(J) +6,000 +9,000 −12,000 −3,000 0 *Q*(J) +10,000 +9,000 −20,000 −3,000 −4,000 *W*(J) −4,000 0 +8,000 0 +4,000

Let's add some of what we've learned about this cycle to our graph.

This cycle is more like a refrigerator than an engine. The net work done was positive, which means work was being done

*on*the gas. The net heat was positive, which means more heat was released to the environment than was generated by doing work on the gas. That heat came from something and was deposited into the environment. This is what a refrigerator does. It uses work to extract heat from something and then deposits that heat outside itself.

### practice problem 2

#### solution

Answer it.

### practice problem 3

#### solution

Answer it.

### practice problem 4

#### solution

Answer it.