## p 240 Self Test 7.1B

Ikeda2M
Posts: 4
Joined: Thu Dec 11, 2014 3:00 am

### p 240 Self Test 7.1B

Can someone please show me step-by-step on how to do the calculations for this problem. I googled the same question but I did not understand the conversions that took place. Thank you for your help Chemistry Community.

Sincerely,

Ikeda

Neil DSilva 1L
Posts: 70
Joined: Fri Sep 26, 2014 2:02 pm

### Re: p 240 Self Test 7.1B

The gases in the four cylinders of an automobile engine expand from 0.22 L to 2.2 L during one ignition cycle. Assuming that the gear train maintains a steady pressure of 9.60 atm on the gases, how much work can the engine do in one cycle?

You can follow the same steps as Sample 7.1 on the previous page.

Anticipate: Work is done by the system, therefore w and $\Delta U$ to be negative (the system loses energy).
PLAN: Use Eq. 3 (from the textbook, pg. 239) to find w and convert L.atm to joules.
Given: $V_{i} = 0.22 L, V_{f} = 2.2 L, P_{ex}=9.6 atm$

Equation 3: $w= -P_{ex}\Delta V$
$w= -(9.6atm)(2.2L-0.22L)$
$w= -19.008 L\cdot atm$
$=-19.008 L\cdot atm \times \frac{101.325J}{1L\cdot atm}= -1925.9856 J$
$= -1.9 kJ$

1.9 kJ of work is done by one ignition cycle.

The only real conversion in this problem is from L.atm to joules and the relationship between these two can be found in the textbook on page 239. Essentially, liters are converted to meters cubed (m3)and atmospheres are converted to pascals (Pa = kg.m-1.s-2) and when these are multiplied, you get the equivalent in joules (joules = kg.m2.s-2).

Ikeda2M
Posts: 4
Joined: Thu Dec 11, 2014 3:00 am

### Re: p 240 Self Test 7.1B

Thank you very much!

Niharika Reddy 1D
Posts: 127
Joined: Fri Sep 26, 2014 2:02 pm

### Re: p 240 Self Test 7.1B

The problem states: the gases in the four cylinders of an automobile engine expand from 0.22 L to 2.2 L during one ignition cycle. Assuming that the gear train maintains a steady pressure of 9.60 atm on the gases, how much work can the engine do in one cycle?

To solve this problem, use the equation w=-PΔV.

We know the constant external pressure is 9.60 atm, and ΔV can be calculated by subtracting the initial volume from the final volume: ΔV=Vfinal-Vinitial=2.2L - 0.22L.

Plugging these values into w=-PΔV, we get the amount of work in L●atm but the unit of work is energy in joules, so we must convert from L●atm to J.

1 L●atm=10-3m3 x 101325 Pa = 101.325 Pa●m3 = 101.325 J

Thus, convert from L●atm to J using the conversion factor 1 L●atm=101.325 J, to get the final answer:

w=-PΔV = -(9.60 atm)(2.2L - 0.22L) = -19 L●atm

Converting to joules, we get:

w= (-19 L●atm)(101.325 J/L●atm) = -1926 J = -1.9 kJ

Someone replied while I was typing my response, but thought I would still post it anyways. Hope this helps!

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