The isobars of one mole of an ideal gas were obtained at three different pressures $(p_1, p_2$ and $p_3)$. The slopes of these isobars are $m_1, m_2$ and $m_3$ respectively. If $p_1<p_2<p_3$, then the correct relation of the slopes is
Show Hint
For an ideal gas, an isobar is a graph of Volume (V) vs. Temperature (T) at constant pressure (P). The graph is a straight line through the origin ($V = (R/P)T$), and the slope is $R/P$. Remember that the slope of this graph is inversely proportional to the pressure.
Step 1: Write the equation of state for an ideal gas.
The ideal gas law is $PV = nRT$.
We are given that $n=1$ mole, so $PV = RT$.
Step 2: Express the relationship for an isobar.
An isobar is a graph plotted at constant pressure ($P$). The graph is plotted with Volume ($V$) on the y-axis and Temperature ($T$) on the x-axis.
We rearrange the ideal gas law to express $V$ as a function of $T$:
\[
V = \left(\frac{R}{P}\right) T.
\]
Step 3: Determine the slope of the isobar.
The equation $V = \left(\frac{R}{P}\right) T$ is a straight line equation of the form $y = m x$, where $y=V$, $x=T$, and $m$ is the slope.
The slope of the isobar is:
\[
m = \frac{R}{P}.
\]
Step 4: Relate the slopes to the given pressures.
We are given three different pressures, $p_1, p_2, p_3$, and their corresponding slopes $m_1, m_2, m_3$.
\[
m_1 = \frac{R}{p_1}, m_2 = \frac{R}{p_2}, m_3 = \frac{R}{p_3}.
\]
Step 5: Determine the relationship between the slopes.
We are given the pressure relationship $p_1<p_2<p_3$.
Since the slope $m$ is inversely proportional to the pressure $P$ (i.e., $m \propto \frac{1}{P}$), a smaller pressure corresponds to a larger slope.
Therefore, the order of the slopes must be the inverse of the order of the pressures.
\[
p_1<p_2<p_3 \implies \frac{1}{p_1}>\frac{1}{p_2}>\frac{1}{p_3}.
\]
Multiplying by $R$ (which is a positive constant) maintains the inequality:
\[
\frac{R}{p_1}>\frac{R}{p_2}>\frac{R}{p_3} \implies m_1>m_2>m_3.
\]
