Question
When $2 \mathrm{~g}$ of a gas $A$ is introduced into an evaluated flask kept at $25^{\circ} \mathrm{C}$, the pressure is found to be one atmosphere. If $3 \mathrm{~g}$ of another gas $B$ is then added to the same flask, the total pressure becomes $1.5$ atm. Assuming ideal gas behaviour, calculate the ratio of the molecular weights $M_{A}: M_{B}$
Step 1
Let's denote the molar mass of gas A as $M_A$. The number of moles of gas A, $n_A$, can be calculated using the formula $n = \frac{m}{M}$, where $m$ is the mass of the gas. So, we have $n_A = \frac{2}{M_A}$. Show more…
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Key Concepts
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A $0.5 \mathrm{dm}^{3}$ flask contains gas $\mathrm{A}$ and $1 \mathrm{dm}^{3}$ flask contains gas $\mathrm{B}$ at the same temperature. If density of $\mathrm{A}$ is twice that of $\mathrm{B}$ and the molar mass of $\mathrm{A}$ is half of $\mathrm{B}$, then the ratio of pressure exerted by gases is (a) $P_{A} / P_{B}=2$ (b) $P_{A} / P_{B}=1$ (c) $P_{A} / P_{B}=4$ (d) $P_{\Lambda} / P_{B}=3$
Gaseous State
Exercises I
Pressure of $1 \mathrm{~g}$ of an ideal gas $\mathrm{A}$ at $27^{\circ} \mathrm{C}$ is found to be 2 bar. When $2 \mathrm{~g}$ of another ideal gas $\mathrm{B}$ is introduced in the same flask at same temperature the pressure becomes 3 bar. What is the relationship between their molecular masses? (a) $2 M_{\mathrm{B}}=M_{\mathrm{A}}$ (b) $M_{\mathrm{B}}=M_{\mathrm{A}}$ (c) $M_{\mathrm{B}}=4 M_{\mathrm{A}}$ (d) $M_{\mathrm{B}}=2 M_{\mathrm{A}}$
Pressure of 1 g of an ideal gas $A$ at $27^{\circ} \mathrm{C}$ is found to be 2 bar. When $2 \mathrm{~g}$ of another ideal gas $\mathrm{B}$ is introduced in the same flask at same temperature the pressure becomes 3 bar. Find a relationship between their molecular masses.
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