For same mass of two different ideal gases of molecular weights $M_{1}$ and $M_{2}$, Plots of log $V$ vs log $p$ at a given constant temperature are shown. Identify the correct option.

• Correct answer
  A

  $M_{1}>M_{2}$
• B

  $M_{1}=M_{2}$
• C

  $M_{1} < M_{2}$
• D

  Can be predicted only if temperature is known

Solution

For ideal gas, $\log \frac{k}{M_{2}}, \log \frac{k}{M_{1}}$ $\because$ $p V=\frac{W}{M} R T$ Let, $W R T=K$ Taking log of both the sides of Eq. (i) $$ \log p+\log V=\log \frac{K}{M} $$ or $\log V=-\log p+\log \frac{K}{M}$ [On compairing Eq. (ii) with $y=m x+C]$ $\log \frac{K}{M_{2}}>\log \frac{K}{M_{1}}$ or $M_{1}>M_{2}$

Hence, the correct answer is option: 'A'.