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thermodynamics GATE

 THERMODYNAMICS

Yes — the confusing part is why $f'(y)=0$. Let's go through that step very slowly.

We started with:

$$dB=(2xy+3)\,dx+x^2\,dy$$

So

$$M=2xy+3,\qquad N=x^2$$

Step 1: Integrate $M$ with respect to $x$

We want to find $B$. Since

$$M=\frac{\partial B}{\partial x}$$

we integrate:

$$B=\int M\,dx$$

Therefore,

$$B=\int(2xy+3)\,dx$$

Here $y$ is treated as a constant, because we're integrating with respect to $x$.

So:

$$\int2xy\,dx=y\int2x\,dx=x^2y$$

and

$$\int3\,dx=3x$$

Thus:

$$\boxed{B=x^2y+3x+f(y)}$$

Why $f(y)$?

This is the important bit.

When you integrate with respect to $x$, you can have any function of $y$ as the "constant."

For example:

$$\frac{\partial}{\partial x}(x^2y+3x+y^5) =2xy+3$$

The $y^5$ disappears because we're differentiating with respect to $x$.

So we don't know yet what that function is. We call it:

$$f(y)$$

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Step 2: Differentiate this with respect to $y$

Now:

$$B=x^2y+3x+f(y)$$

Differentiate with respect to $y$:

$$\frac{\partial B}{\partial y} = \frac{\partial}{\partial y}(x^2y) + \frac{\partial}{\partial y}(3x) + \frac{\partial}{\partial y}f(y)$$

Therefore:

$$\frac{\partial B}{\partial y} = x^2+0+f'(y)$$

so:

$$\boxed{\frac{\partial B}{\partial y}=x^2+f'(y)}$$

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Step 3: Now use the information we already have

From our original differential:

$$dB=(2xy+3)dx+x^2dy$$

we know that the coefficient of $dy$ is

$$\boxed{N=x^2}$$

But by definition,

$$N=\frac{\partial B}{\partial y}$$

Therefore:

$$x^2+f'(y)=x^2$$

Subtract $x^2$ from both sides:

$$\boxed{f'(y)=0}$$

And if

$$f'(y)=0$$

then $f(y)$ must simply be a constant:

$$f(y)=C$$

Hence:

$$\boxed{B=x^2y+3x+C}$$

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The key idea

Think of it like this:

$$\boxed{\text{Integrating with respect to }x \Rightarrow \text{constant can actually be any function of }y}$$

That's why we initially write

$$\boxed{B=x^2y+3x+f(y)}$$

rather than simply $+C$.

Then the second piece of information, $N=x^2$, tells us what $f(y)$ must be.

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And this is exactly what happens with thermodynamic properties

Suppose you know

$$dU=T\,dS-P\,dV$$

and you know $T(S,V)$. You could integrate $T\,dS$ with respect to $S$:

$$U=\int T(S,V)\,dS+f(V)$$

Notice that the "constant" is actually a function of $V$ because you integrated with respect to $S$.

Then you use the $-P\,dV$ part to determine $f(V)$.

That's the same mathematical procedure you're seeing in this example.