The First Law of Thermodynamics

rputra

I am working on this problem:

Q: During an isothermal expansion, a confined ideal gas does -150 J of work against its surroundings. which of the following describes the heat transfer during this process? (a) 150 J of heat was added to the gas; (b) 150 J of heat was removed from the gas; (c) 300 J of heat was added to the gas; (d) 300 J of heat was removed from the gas.

The answer is (a). The explanation on the book is that "... since ΔU = Q + W by the first law of thermodynamics, and since ΔU = 0 due to isothermal process, therefore Q = -W. Since W = -150 J, it must be true that Q = +150 J."

My question is that all textbooks write the law as ΔQ = ΔU + W, why does this book writes it differently? Any help would be very much appreciated. Thanks for your time.

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Chestermiller

Mentor
I am working on this problem:

Q: During an isothermal expansion, a confined ideal gas does -150 J of work against its surroundings. which of the following describes the heat transfer during this process? (a) 150 J of heat was added to the gas; (b) 150 J of heat was removed from the gas; (c) 300 J of heat was added to the gas; (d) 300 J of heat was removed from the gas.

The answer is (a). The explanation on the book is that "... since ΔU = Q + W by the first law of thermodynamics, and since ΔU = 0 due to isothermal process, therefore Q = -W. Since W = -150 J, it must be true that Q = +150 J."

My question is that all textbooks write the law as ΔQ = ΔU + W, why does this book writes it differently? Any help would be very much appreciated. Thanks for your time.
Hi rputra. Welcome to physics forums.

Some books write the first law as Q-W=ΔU, where W is the work done by the system on the surroundings. Other books write the first law as Q+W=ΔU, where W is the work done by the surroundings on the system. You just need to determine which basis is being used by the particular book.

Chet

George Jones

Staff Emeritus
Gold Member
Adding specific examples of what Chestermiller wrote:
Because heat and work are things transferred rather than contained, there's a bit of ambiguity with regard to the sign convention. With heat it just makes sense to define heat as positive if it makes the object gain energy, negative if it makes the object lose energy. With work, it's not so obvious. Some define work as positive if the object performs work on the environment, others define it exactly the opposite. The first law of thermodynamics is $\Delta U = Q-W$ with the first convention but is $\Delta U = Q+W$ with the latter convention.
I would like to add, that from my experiences, $\Delta U = Q-W$ is a more commonly used convention.
In physics we use ΔU=Q-W with signs.

Work done by system is positive and work done on system is negative.

However in chemistry,

ΔU=Q+W

Work done by system is negative and work done on system is positive.

Just difference in conventions.
These days, the $\Delta U = Q + W$ convention is used quite often in physics texts. For example, the text "Thermal Physics" by Schroeder is widely used in North America and it uses this convention. This convention also is used in the first-year physics by Serway and by Knight. Over the last couple of years, I have used all of these as texts for physics courses that I have taught.

The first-year book by Halliday, Resniick, and Walker, however, uses the other convention. So, both conventions are used extensively, and students should be careful.

At my university students see thermal/statistical physics in first-year in either Knight or Serway (+W), in second-year in Schroeder (+W), and in fourth-year in Baierlein (-W).

rputra

Thank you all for your generous responses. Since this is my first use, I am wondering if there is any way of selecting one of them as the best answer, sort of rewarding your time and effort. Thank you again for your time.

Chestermiller

Mentor
Thank you all for your generous responses. Since this is my first use, I am wondering if there is any way of selecting one of them as the best answer, sort of rewarding your time and effort. Thank you again for your time.
If you want to do something like that, just go to the response (or responses) you like best and click the Thanks button on each of them.

Chet