The isotherms of an ideal gas at $T_1, T_2, T_3$ along with their slopes (m) (in the brackets) are shown…
The isotherms of an ideal gas at $T_1, T_2, T_3$ along with their slopes (m) (in the brackets) are shown here. If $T_1>T_2>T_3$, then the correct order of slopes of these isotherms is
$m_2>m_1>m_3$
$m_3>m_2>m_1$
$\mathrm{m}_2>\mathrm{m}_3>\mathrm{m}_1$
$\mathrm{m}_1>\mathrm{m}_2>\mathrm{m}_3$
Solution
$\mathrm{P}=\mathrm{nRT} \times \frac{1}{\mathrm{~V}}(\mathrm{y}=\mathrm{mx}+\mathrm{c}$ where $\mathrm{c}=0)$
Thus, a higher value of $\mathrm{T}$ would increase the value of the slope of the $\mathrm{P}$ vs. $\frac{1}{\mathrm{~V}}$ curve.
Therefore, the correct order of slopes would be:$\mathrm{m}_1>\mathrm{m}_2>\mathrm{m}_3$
