January 23, 2021, 12:33:11 PM
Forum Rules: Read This Before Posting


Topic: Work Questions  (Read 18213 times)

0 Members and 1 Guest are viewing this topic.

Offline govibe

  • Regular Member
  • ***
  • Posts: 59
  • Mole Snacks: +1/-4
Work Questions
« on: October 19, 2007, 05:45:52 PM »
"Consider the following changes.

    (a) COCl2(g) CO(g) + Cl2(g)
    (b) N2(g) N2(l)
    (c) CO(g) + H2O(g) H2(g) + CO2(g)
    (d) Ca3P2(s) + 6 H2O(l) 3 Ca(OH)2(s) + 2 PH3(g)
    (e) 2 CH3OH(l) + 3 O2(g) 2 CO2(g) + 4 H2O(l)
    (f) I2(s) I2(g)

1. At constant temperature and pressure, in which of these changes is work done by the system on the surroundings? (Select all that apply.)

2. At constant temperature and pressure, in which of these changes is work done by the surroundings on the system?

3.In which of them is no work done?"

I'm confused what they mean by work done on the system... does that mean it's exothermic? How do I figure out for each one whether the work is done on the surrounding or by the surrounding? *delete me*

Offline Padfoot

  • Full Member
  • ****
  • Posts: 221
  • Mole Snacks: +23/-2
Re: Work Questions
« Reply #1 on: October 19, 2007, 06:48:06 PM »
Energy can be transferred in 2 ways - heat and work.
Work has been done by the system if volume increases.

PV=nRT
V=nRT/P
What will happen if P,T are constant but n changes?

Offline govibe

  • Regular Member
  • ***
  • Posts: 59
  • Mole Snacks: +1/-4
Re: Work Questions
« Reply #2 on: October 19, 2007, 07:11:20 PM »
Volume increases... But I still don't see how that is involved in the problem...

Offline Padfoot

  • Full Member
  • ****
  • Posts: 221
  • Mole Snacks: +23/-2
Re: Work Questions
« Reply #3 on: October 19, 2007, 07:34:10 PM »
Volume increases... But I still don't see how that is involved in the problem...

Work has been done by the system if volume increases.
Work= P*change V

For a), I assume an arrow was meant after COCl2(g), you therefore have more mol of gas on the products side than reactants side.  Given P and T are constant, the ideal gas law tells us that V increases and hence work is done by the system.
Note that if V decreases, work is done by the surroundings on the system.

We are only taking into account moles of gas for this by the way - do not count (s) or (l) as they won't influence the pressure.

Edit: (l) instead of (g)
« Last Edit: October 19, 2007, 07:46:08 PM by Padfoot »

Offline govibe

  • Regular Member
  • ***
  • Posts: 59
  • Mole Snacks: +1/-4
Re: Work Questions
« Reply #4 on: October 19, 2007, 07:45:49 PM »
For the no work done question, I checked b,c, and f but didn't get it right...

Offline Padfoot

  • Full Member
  • ****
  • Posts: 221
  • Mole Snacks: +23/-2
Re: Work Questions
« Reply #5 on: October 19, 2007, 07:54:31 PM »
For the no work done question, I checked b,c, and f but didn't get it right...
c) is the only one in which no work is done (in my opinion).
(since it is the only eqn with equal mol of gas on both sides)

Did you remember coefficients and the states of reactants and products?

Offline govibe

  • Regular Member
  • ***
  • Posts: 59
  • Mole Snacks: +1/-4
Re: Work Questions
« Reply #6 on: October 19, 2007, 09:45:40 PM »
wait, so how does states effect the results? I thought only the number of moles on each side did...

Offline Padfoot

  • Full Member
  • ****
  • Posts: 221
  • Mole Snacks: +23/-2
Re: Work Questions
« Reply #7 on: October 19, 2007, 09:50:52 PM »
wait, so how does states effect the results? I thought only the number of moles on each side did...

We are only taking into account moles of gas for this by the way - do not count (s) or (l) as they won't influence the pressure.

Offline govibe

  • Regular Member
  • ***
  • Posts: 59
  • Mole Snacks: +1/-4
Re: Work Questions
« Reply #8 on: October 21, 2007, 06:04:02 PM »
^If it's only number of moles, isn't I2(s) -> I2(g) have the same number of moles on both sides? Then why isn't there any work done?

Offline Padfoot

  • Full Member
  • ****
  • Posts: 221
  • Mole Snacks: +23/-2
Re: Work Questions
« Reply #9 on: October 21, 2007, 09:10:40 PM »
We are only taking into account moles of gas for this by the way - do not count (s) or (l) as they won't influence the pressure.

I2(s) -> I2(g) 

In the example above there are more moles of gas on the right (1 on right, none on the left).  Since 'n' is greater (and P,T constant) volume increases - work is done by the system.
 

Sponsored Links