Which of the following plot(s) is/are correct representation(s) of Boyle's Law?
Show Hint
Always formulate the explicit mathematical equation for each axis variable before verifying the graph shape:
- $y \propto x \rightarrow$ straight line through origin.
- $y \propto \frac{1}{x} \rightarrow$ rectangular hyperbola.
- $\log y = - \log x + c \rightarrow$ straight line with slope $-1$.
Step 1: Understanding the Question:
We need to determine which of the given plots correctly represent Boyle's Law mathematically.
Step 2: Key Formula or Approach:
Boyle's Law states that for a fixed amount of gas at constant temperature, the pressure ($P$) is inversely proportional to its volume ($V$):
\[ P \propto \frac{1}{V} \implies P = \frac{k}{V} \implies PV = k \text{ (where } k \text{ is a constant)} \] Step 3: Detailed Explanation:
Let us evaluate each of the plots shown in the options:
- Plot (A): $P$ vs $\frac{1{V}$:
Since $P = k \left(\frac{1}{V}\right)$, this is of the linear form $y = mx$ where $y = P$ and $x = \frac{1}{V}$ with a positive slope $m = k$.
This plot is a straight line passing through the origin. Thus, Plot (A) is correct.
- Plot (B): $\log P$ vs $\log V$:
Taking the logarithm of both sides of $PV = k$:
\[ \log(PV) = \log k \implies \log P + \log V = \log k \]
\[ \log P = -\log V + \log k \]
This is of the linear form $y = mx + c$ with slope $m = -1$ and a positive intercept $c = \log k$.
This plot is a straight line with a negative slope. Thus, Plot (B) is correct.
- Plot (C): $P$ vs $V$:
Since $P$ and $V$ are inversely proportional, as volume increases, pressure decreases non-linearly.
This plot is a rectangular hyperbola. Thus, Plot (C) is correct.
- Plot (D): $P$ vs $V$ showing a straight line with a negative slope:
This represents a linear relationship $P = -mV + c$, which is mathematically incorrect for Boyle's law. Thus, Plot (D) is incorrect.
Step 4: Final Answer:
The correct options are (A), (B), and (C).
