Q6 / 160 · TS EAMCET 2023 (13 May Shift 2) · Chemistry

$4 \mathrm{~g}$ of an ideal gas $\mathrm{A}$ (molar mass $=\mathrm{M}_{\mathrm{A}}$ ) present in a vessel of a volume $\mathrm{V}$ litre exerted a pressure of $5 \mathrm{~atm}$ at $300 \mathrm{~K}$. When $16 \mathrm{~g}$ of another ideal gas B (molar mass = MB) was introduced into this vessel at the same temperature, its increased to $10 \mathrm{~atm}$. What is the correct relationship between $\mathrm{M}_{\mathrm{A}}$ and $\mathrm{M}_{\mathrm{B}}$ ?

  • A
    $\mathrm{M}_{\mathrm{A}}=4 \mathrm{MB}$
  • B
    $\mathrm{MA}=\mathrm{MB}$
  • C
    $\mathrm{MA}=2 \mathrm{M}_{\mathrm{B}}$
  • Correct answer
    D
    $4 \mathrm{M}_{\mathrm{A}}=\mathrm{M}_{\mathrm{B}}$

Solution

$\mathrm{PV}=\mathrm{n} \mathrm{R}$ T (ideal gas equation) $$ \mathrm{PV}=\frac{\mathrm{wRT}}{\mathrm{M}} $$ $\therefore \mathrm{M} \alpha \mathrm{w}$ and $\mathrm{M} \alpha \frac{1}{\mathrm{PV}}$ $$ \begin{aligned} & \therefore \frac{M_A}{M_B}=\frac{w_A}{P_A V} \times \frac{P_B V}{w_B} \\ & \frac{M_A}{M_B}=\frac{4}{5 V} \times \frac{5 V}{16}\left\{P_B=10-5=5\right\} \\ & \therefore \quad 4 M_A=M_B \end{aligned} $$

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

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