
A polyethene bag 3 litre capacity is partially filled by 1 litre of helium gas at 0.3 atm at 300 K. Subsequently, enough neon gas is filled to make the total pressure 0.4 atm at 300 K. Calculate ratio of moles of Ne to He in the container.
(A) 1/4
(B) 1/2
(C) 1/3
(D) 3/4
Answer
484.5k+ views
Hint: Number of moles of two different gases when their volumes are given can be known from Avogadro’s law. Avogadro’s law is one of the gas laws given for ideal gas.
Complete answer:
According to Avogadro’s law under similar conditions of temperature and similar pressure, equal volumes of gases contain equal numbers of moles. The formula for calculating mole ratios from Avogadro law is given by,
$\dfrac{{{V}_{1}}}{{{n}_{1}}}=\dfrac{{{V}_{2}}}{{{n}_{2}}}$
Given initial volume $\left( {{V}_{1}} \right)$ of helium gas is 1 litre but the volume of neon gas $\left( {{V}_{2}} \right)$ has to be found.
- As the pressures of two gases are given, the volume of second gas can be known from Boyle’s law.
Boyle’s law states that pressure and volume of a gas are inversely related to each other which means that the pressure of the gas increases with decrease in volume and vice-versa.
From the formula of Boyle’s law, let us calculate the volume of the second gas,
${{P}_{1}}{{V}_{1}} = {{P}_{2}}{{V}_{2}}$
$0.3\times 1= 0.4\times {{V}_{2}}$
${{V}_{2}}=\dfrac{0.3\times 1}{0.4}$
${{V}_{2}} = 0.75$
The volume of the second gas Neon is 0.75 litre. Upon substituting the volumes of the two gases in the Avogadro’s gas law, we get,
$\dfrac{{{V}_{1}}}{{{n}_{1}}}=\dfrac{{{V}_{2}}}{{{n}_{2}}}$
$\dfrac{{{V}_{1}}}{{{V}_{2}}}=\dfrac{{{n}_{1}}}{{{n}_{2}}}$
$\dfrac{1}{0.75}=\dfrac{1}{\dfrac{3}{4}}=\dfrac{{{n}_{1}}}{{{n}_{2}}}$
Therefore from the above value the ratio of number of moles of Ne : He ,the correct option is option (D) 3/4.
So, the correct answer is “Option D”.
Note: The ratio of $n_1}$ and $n_2$ is 4:3. But the ${{n}_{1}}$, we considered is Helium and ${{n}_{2}}$ is Neon. The given question is asking to calculate the ratio of moles of Neon and Helium so the answer for the above question is 3/4.
Complete answer:
According to Avogadro’s law under similar conditions of temperature and similar pressure, equal volumes of gases contain equal numbers of moles. The formula for calculating mole ratios from Avogadro law is given by,
$\dfrac{{{V}_{1}}}{{{n}_{1}}}=\dfrac{{{V}_{2}}}{{{n}_{2}}}$
Given initial volume $\left( {{V}_{1}} \right)$ of helium gas is 1 litre but the volume of neon gas $\left( {{V}_{2}} \right)$ has to be found.
- As the pressures of two gases are given, the volume of second gas can be known from Boyle’s law.
Boyle’s law states that pressure and volume of a gas are inversely related to each other which means that the pressure of the gas increases with decrease in volume and vice-versa.
From the formula of Boyle’s law, let us calculate the volume of the second gas,
${{P}_{1}}{{V}_{1}} = {{P}_{2}}{{V}_{2}}$
$0.3\times 1= 0.4\times {{V}_{2}}$
${{V}_{2}}=\dfrac{0.3\times 1}{0.4}$
${{V}_{2}} = 0.75$
The volume of the second gas Neon is 0.75 litre. Upon substituting the volumes of the two gases in the Avogadro’s gas law, we get,
$\dfrac{{{V}_{1}}}{{{n}_{1}}}=\dfrac{{{V}_{2}}}{{{n}_{2}}}$
$\dfrac{{{V}_{1}}}{{{V}_{2}}}=\dfrac{{{n}_{1}}}{{{n}_{2}}}$
$\dfrac{1}{0.75}=\dfrac{1}{\dfrac{3}{4}}=\dfrac{{{n}_{1}}}{{{n}_{2}}}$
Therefore from the above value the ratio of number of moles of Ne : He ,the correct option is option (D) 3/4.
So, the correct answer is “Option D”.
Note: The ratio of $n_1}$ and $n_2$ is 4:3. But the ${{n}_{1}}$, we considered is Helium and ${{n}_{2}}$ is Neon. The given question is asking to calculate the ratio of moles of Neon and Helium so the answer for the above question is 3/4.
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