How is value of slope $= 1$ here?

Question

So , I am going to represent equation of Boyle's law. I will provide the math . It’s ok if you don’t know chemistry.

So , $PV = nRT$ ( Ideal gas equation ) $y= p$ , $x=V$ and $nRT = K$.

$Yx = K$ Therefore , graph of hyperbola we got.

Let us do $Y= K/x$

Now , Putting $\log$ on both sides.

$\log P = \log K + \log(1/V)$

Now , assuming we can say that

$P = K + x$ here.

So , comparing it with $y= mx+c$.

$m(x) + K = P$.

So , my Q is that why did we assume slope or $m = 1$ here ? A possibility is also that $x= 4$ and $m= 1/4$ .

Edit Q of someone: why I took $\log$ on both sides

I need to solve this Q part b and c.

Answer 1

Let's take $PV=nRT$.

For option (b) and (c), we want the graph of $\log P$ and $\log(-V)$, so let's take the logarithm on both sides considering that $n$ and $T$ are constants.

\begin{align*} \log(PV)& =\log(nRT) \\ \log P+\log V&=\log k \\ \log P&=-\log V+\log k \\ \log P &= \log(V^{-1})+\log k \end{align*}

Now, we can see that the slope of $\log P$ v/s $\log V$ graph is $1$. Therefore, (b) is the correct answer.

Answer 2

Let's work out by cases. Boyle's law is stated similar to the statement below:

If the temperature $T$ is constant, the pressure $P$ is inversely proportional to the volume $V$.

This means that $PV = T$.

Plotting $V$ v/s $P$, we see that the graph is $(a)$. We then eliminate this option.

Similarly, if we plot $PV$ v/s $V$, we see that the graph will be a horizontal line as $y$is constant for all $V$. We then eliminate $(d)$ from the options.

Lastly, we take the logarithm on both side as follows: \begin{align*}PV &= T \\ \log P + \log V &= \log T \\ \log P &= - \log V + \log T\\ \log P &= \log(V^{-1}) + \log T \end{align*}

We then make the scale of $y$ and $x$ to be $\log P$ and $\log(V^{-1})$. By rewriting $\log(V^{-1})$, we get $-\log V$. Let $\log T \to T$. Then, we can rewrite the equation to be $$y = -x + T.$$

We see that this equation represents $(c)$. Therefore, $(b)$ must be the answer.