1
$\begingroup$

In the equation, $\Delta E = q - W$, according to my book, $W$ is positive when it is done by the system and negative when it is done on the system.

But in this problem:

$\pu{3600 kJ}$ heat is entered in a gas cylinder, calculate the value of $∆E$ if the volume is not constant and work of $\pu{-800 kJ}$ was performed by the gas.

the sign is different. How should I make the equation?

Improve this question
$\endgroup$
4
  • $\begingroup$ Who says the sign is different? $\endgroup$
    – Ivan Neretin
    May 4, 2018 at 21:02
  • $\begingroup$ Here the work was done by the system but it is stated as negative. $\endgroup$
    – Hammad Ahmed
    May 4, 2018 at 21:21
  • $\begingroup$ All that means is that the surroundings in this problem are doing work on the system. Have you never worked with negative numbers? $\endgroup$
    – Chet Miller
    May 4, 2018 at 22:39
  • 3
    $\begingroup$ @ChesterMiller is "Have you never worked with negative numbers?" really necessary? Please be nice to users. Thanks! $\endgroup$
    – JSCoder says Reinstate Monica
    May 5, 2018 at 1:10

2 Answers 2

Reset to default
1
$\begingroup$

I always got lost in remembering the rules for the signs, so I just ignore them and think about the two parts from the standpoint of the system (cylinder):

If heat enters the system, then it gets hotter so that contribution is positive. This part's fairly intuitive.

If I let the cylinder expand, it's relaxing to a less compressed state and so loses energy, which means the work contribution is negative. In order to compress the cylinder I have to push on it, adding some of my own energy to the system, so the work contribution is positive.

In your case, the change in energy from the heating (positive), and loss of energy through expansion (negative) is: $$ \Delta E = 3600kJ - 800kJ $$

Edit: The problem itself is ambiguous regarding whether it's expansion of contraction. My answer assumes expansion since that's what would naturally happen upon heating a gas cylinder. Gaurang's answer is correct if you read it as contraction.

Improve this answer
$\endgroup$
5
  • $\begingroup$ I am not sure if I understood the main part of your answer. That said though, our final values are contradictory. Please have a look at my answer below. $\endgroup$
    – Gaurang Tandon
    May 5, 2018 at 6:26
  • $\begingroup$ We're both saying the same thing, but we're interpreting the question differently: expansion vs contraction. I assumed expansion because if you added heat to a cylinder the gas would naturally expand. $\endgroup$
    – Mark Wolfman
    May 5, 2018 at 6:49
  • $\begingroup$ Since in chemistry, work $=-P\Delta V$, that means there definitely is an expansion in the given question (since work given is negative). But, you seem to imply that I assumed contraction of volume. Though, I don't know why. I never said in my answer that I'm assuming contraction. In fact, I didn't even mention volume changes at all! ;) $\endgroup$
    – Gaurang Tandon
    May 5, 2018 at 7:03
  • $\begingroup$ Oh, so we agree it's expansion then? A compressed piston has a higher potential energy than an expanded one, so the when the cylinder expands it has to lose energy. The only way this works is if the $w_{PV}$ winds up being subtracted from the heat. $\endgroup$
    – Mark Wolfman
    May 5, 2018 at 7:21
  • 1
    $\begingroup$ No, sorry. With that last comment I ended up contradicting myself. You're right, though, I was assuming contraction. This problem is indeed ambiguous. $\endgroup$
    – Gaurang Tandon
    May 5, 2018 at 7:29
0
$\begingroup$

You can interpret the process of work in the reverse direction. A work of $\pu{-800kJ}$ is done by the gas, that implies $\pu{800kJ}$ work is done on the gas.

Now, since positive work is done on the gas, positive energy would be supplied to the gas. Recall that work is essentially the transfer of energy. So, here the internal energy of the gas would increase.

Thus, final $\Delta E$ would be $(\pu{+3600kJ}) + (\pu{+800kJ})=\pu{+4400kJ}$.

Improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.