How do I tell if any quantity is a state function when heat and work depend on the path?

Amma says I study best at 5 am with filter coffee, so here I am with thermodynamics, and the very first page of it is eating me.

NCERT says internal energy $U$ is a state function while heat $q$ and work $w$ are not. I can repeat it, but a test question asked "which of the following are state functions: $q$ , $w$ , $q + w$ , $H$ , $pV$ , $S$ " and I guessed. Is there a rule I can apply to any quantity, or do I just memorise the list?

My real confusion is the isothermal expansion of an ideal gas from $V_1$ to $V_2$ . I read that you can do it reversibly or irreversibly. In both cases the initial and final states are the same, so $\Delta U$ is the same (zero). Then work should also be the same? But the book says work done differs. How can $q$ and $w$ individually change but their sum doesn't?

Also, I am always unsure about the sign convention. Is expansion work positive or negative in NCERT? Last year I lost marks on exactly this, so I want to settle it now.

7 answers & replies

[+]Neel Chatterjee12thSample· 1mo ago

A state function depends only on the present state of the system (its $p$ , $V$ , $T$ , composition) and not on how it got there. So $\Delta$ (state function) depends only on initial and final states. Test: go around a cycle and come back to the start. A state function returns to its original value, so its change over a cycle is zero. $q$ and $w$ don't, that's why they are path functions.

For your list: $q$ no, $w$ no, $q+w$ yes (it equals $\Delta U$ ), $H$ yes, $pV$ yes (all three variables are state properties), $S$ yes. The first law $\Delta U = q + w$ is a statement that two path-dependent things add up to a path-independent one.

[+][–]Kavya NaikDropperSample· 1mo agoAuthor
the cycle test is going on a sticky note on my table. thank you neel!

[+]Yash Tiwari11thSample· 1mo ago

bhai, isothermal mein dono cases mein initial aur final state same hai, ΔU same hai, to work bhi same hoga na? I think work is the same for reversible and irreversible isothermal expansion

[+]Arjun Nair12thSample· 1mo ago

Yash, work isn't set by the end states, it's set by the path. ΔU is the same (zero for an ideal gas at constant T) so $q = -w$ in both, but the work itself differs.

Reversible isothermal: $w_{rev} = -2.303\,nRT\log\dfrac{V_2}{V_1}$ , which is the maximum work the gas can do. Irreversible expansion against a constant external pressure: $w = -p_{ex}(V_2 - V_1)$ , which is smaller in magnitude.

Since $w$ differs and ΔU is the same, $q$ must differ by the same amount. That's the whole point: $q$ and $w$ change with path but their sum doesn't. Free expansion into vacuum is the extreme case, $p_{ex} = 0$ so $w = 0$ .

[+][–]Yash Tiwari11thSample· 1mo ago
oh got it, ΔU same hai lekin w alag. isliye q bhi alag. thanks arjun bhaiya 🙏

[+]Vivek Mishra12thSample· 1mo ago

Extensive and intensive, because they come with state functions in tests:

Extensive (depend on amount): $V$ , $U$ , $H$ , $S$ , $G$ , mass, number of moles, heat capacity. Intensive (do not): $T$ , $p$ , density, refractive index, molar volume, specific heat.

A simple trick: divide an extensive by moles or mass and it becomes intensive. Molar volume is intensive, volume is extensive.

[+]Riya Singh12thSample· 1mo ago

Sign convention (IUPAC, followed by NCERT): work done on the system is positive, work done by the system is negative. So when a gas expands, $w$ is negative. Compression, $w$ is positive. Heat absorbed by the system is positive, heat released is negative. With that, $\Delta U = q + w$ works directly and you don't need to flip signs.

Two more handy lines: at constant volume $q_V = \Delta U$ , at constant pressure $q_p = \Delta H$ .

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