A gas absorbs 45 kJ of heat and does 29 kJ of work. Calculate the change in energy.

The answer is 16 kJ, but I don't understand why.

If a gas absorbs heat and does 29 kJ of work. Wouldn't they both be positive?
Change in energy = q + w
45 kJ + 29 kJ = 74 kJ

Why is the work negative?

Thanks.

Work done BY the system is negative.

Here is the way to remember it.
You know delta H is + when heat is absorbed (receiving) and - when the heat is emitted(giving).
Work is + when receiving) and - when giving.

This is wrong it should be 16 kJ as you subtract 45-29 which equals 16.

Ahhh, that's really helpful. Thanks for clearing it up.

كول خرا

Ah, I see what's going on here. So, when we talk about work done by a gas, it is considered positive if work is being done BY the gas, and negative if work is being done ON the gas. In this case, the gas is doing 29 kJ of work, which means it is releasing energy. So, we consider it negative.

Let's break it down:
Change in energy = Heat (q) + work (w)
Change in energy = 45 kJ - 29 kJ
Change in energy = 16 kJ

So, the change in energy is indeed 16 kJ, but since the gas is doing work, we consider it negative. Hope this clears things up!

The confusion in this case arises from understanding the sign conventions for heat and work in thermodynamics. Let's break down the situation to clarify it.

In thermodynamics, heat (q) and work (w) are considered as energy transfer mechanisms. The signs of q and w depend on whether energy is gained or lost by the system.

For heat:
- Positive heat (q > 0) means heat is gained by the system from the surroundings.
- Negative heat (q < 0) means heat is lost by the system to the surroundings.

For work:
- Positive work (w > 0) means work is done by the system on the surroundings.
- Negative work (w < 0) means work is done on the system by the surroundings.

Now, let's apply this to your example. The gas absorbs 45 kJ of heat. Since the heat is absorbed by the gas, the sign for q is positive, q = +45 kJ.

Next, the gas does 29 kJ of work. In this case, the work is done by the gas on the surroundings. As per the sign convention, the work done by the system is considered positive. Therefore, the sign for w is +29 kJ.

Now, we can calculate the change in energy (ΔE) by summing the heat and work:
ΔE = q + w

Substituting the values:
ΔE = +45 kJ + (+29 kJ)
= +74 kJ

Based on this calculation, the change in energy seems to be +74 kJ, not 16 kJ as you mentioned.

However, it's important to note that the calculation you provided, where the work is considered positive, is a different scenario from the original problem. In the original problem, the work is stated as 29 kJ, not -29 kJ.

If the work was indeed -29 kJ, then the calculation would be as follows:
ΔE = +45 kJ + (-29 kJ)
= +16 kJ

Therefore, the correct answer of 16 kJ is obtained by assuming the work done by the gas is -29 kJ, meaning work is done on the gas by the surroundings.