Chemistry - Ideal Gas Question with Solution | TestHub

This problem requires matching the ideal gas law relationships with their corresponding graphs. The ideal gas law is given by $PV=nRT$, where $n$ and $R$ are constants.

P: $\frac{1}{V^2}$ vs P at constant T and n. From $PV=nRT$, at constant $T$ and $n$, $PV=constant$. This is Boyle's Law. As P increases, V decreases, so $\frac{1}{V^2}$ increases non-linearly. The graph showing an increasing non-linear relationship is (3).

Q: V vs $\frac{1}{T}$ at constant P and n. From $PV=nRT$, at constant $P$ and $n$, $V=\frac{nR}{P}T$. This shows $V\propto T$. However, the graph is $V$ vs $\frac{1}{T}$. The relationship is $V=(nR/P)\times T$. The graph of $V$ vs $T$ is a straight line through the origin. The graph of $V$ vs $1/T$ is a hyperbola that decreases as $1/T$ increases, or increases as $T$ decreases. Graph (4) shows a decreasing curve.

R: log P vs log V for ideal gas. Taking the logarithm of Boyle's Law ($PV=k$):

$$\log (PV)=\log (k)⟹\log P+\log V=\log k⟹\log P=-\log V+\log k$$

This is in the form $y=mx+c$, where $y=\log P$, $x=\log V$, the slope $m=-1$, and the y-intercept $c=\log k$. Graph (1) shows a straight line with a negative slope.

S: V vs $\frac{1}{P^2}$ at constant T and n. From $PV=nRT$, at constant $T$ and $n$, $V=\frac{nRT}{P}$. We want $V$ vs $\frac{1}{P^2}$. Let $x=\frac{1}{P^2}$, so $P=\frac{1}{\sqrt{x}}$. Substituting into the V equation: $V=nRT\sqrt{x}=nRT\sqrt{\frac{1}{P^2}}$. This is a non-linear, increasing curve (square root function). Graph (2) shows a non-linear, increasing curve.