
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
482.7k+ 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.
Recently Updated Pages
Master Class 9 General Knowledge: Engaging Questions & Answers for Success

Earth rotates from West to east ATrue BFalse class 6 social science CBSE

The easternmost longitude of India is A 97circ 25E class 6 social science CBSE

Write the given sentence in the passive voice Ann cant class 6 CBSE

Convert 1 foot into meters A030 meter B03048 meter-class-6-maths-CBSE

What is the LCM of 30 and 40 class 6 maths CBSE

Trending doubts
Which one is a true fish A Jellyfish B Starfish C Dogfish class 11 biology CBSE

What is the difference between superposition and e class 11 physics CBSE

State and prove Bernoullis theorem class 11 physics CBSE

1 ton equals to A 100 kg B 1000 kg C 10 kg D 10000 class 11 physics CBSE

State the laws of reflection of light

One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE
