Sign of work done by the system in first law of thermodynamics equation

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?

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

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.

• 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. – Gaurang Tandon May 5 at 6:26
• 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. – Mark Wolfman May 5 at 6:49
• 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! ;) – Gaurang Tandon May 5 at 7:03
• 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. – Mark Wolfman May 5 at 7:21
• No, sorry. With that last comment I ended up contradicting myself. You're right, though, I was assuming contraction. This problem is indeed ambiguous. – Gaurang Tandon May 5 at 7:29

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 $(\pu{+3600kJ}) + (\pu{+800kJ})=\pu{+4400kJ}$.