
An open vessel containing air is heated from $300K$ to $400K$. The fraction of air, which gives out with respect to originally present is:
A. $\dfrac{3}{4}$
B. $\dfrac{1}{4}$
C. $\dfrac{2}{3}$
D. $\dfrac{1}{8}$
Answer
564k+ views
Hint: You should know that the state of an amount of gas is determined by its pressure, volume and temperature. You can use the ideal gas equation for calculating this question. The temperature used in the equation of state is an absolute temperature who’s appropriate S.I. The unit is the kelvin.
Complete step by step solution:
Let us first know about the ideal gas law.
So, the ideal gas law which is also called the general gas equation, is the equation of state of a hypothetical ideal gas. It is often written in an empirical formula:
$PV=nRT$ where,
P is the pressure,
V is the volume,
n is the amount of substance,
R is the ideal gas constant and
T is the temperature.
This equation is the same for all the gases.
So, here as the vessel is open, the pressure will remain constant. There is no change in the volume of the vessel. Thus, the volume will also remain constant. Since, pressure (P) and volume (V) are constant and R is already a universal constant therefore, the ideal equation can be written as:
$\dfrac{{{n}_{1}}}{{{n}_{2}}}=\dfrac{{{T}_{2}}}{{{T}_{1}}}$
Where,
${{n}_{1}}$ is the initial number of moles,
${{n}_{2}}$ is the final number of moles,
${{T}_{1}}$ is the initial temperature and
${{T}_{2}}$ is the final temperature.
Here, ${{T}_{1}}$ is $300\text{K}$ and ${{T}_{2}}$ is $400\text{K}$ and let’s consider ${{n}_{1}}$ be $1$. So, the fraction of air present in the vessel will be:
$\dfrac{1}{{{n}_{2}}}=\dfrac{{{T}_{2}}}{{{T}_{1}}}$
Then, ${{n}_{2}}=\dfrac{{{T}_{1}}}{{{T}_{2}}}$
Thus, ${{n}_{2}}=\dfrac{300}{400}=\dfrac{3}{4}$
Therefore, the fraction of air present in the vessel is $\dfrac{3}{4}$. Therefore, the gas escaped out will be:
${{n}_{1}}-{{n}_{2}}=1-\dfrac{3}{4}=\dfrac{1}{4}$
Hence, the correct option is B.
Note: Ideal gas law is a good approximation of the behaviour of many gases under many conditions, although it has several limitations. Generally, it is a combination of the empirical Boyle’s law, Charles’ law, Avogadro’s law and Gay-Lussac’s law.
Complete step by step solution:
Let us first know about the ideal gas law.
So, the ideal gas law which is also called the general gas equation, is the equation of state of a hypothetical ideal gas. It is often written in an empirical formula:
$PV=nRT$ where,
P is the pressure,
V is the volume,
n is the amount of substance,
R is the ideal gas constant and
T is the temperature.
This equation is the same for all the gases.
So, here as the vessel is open, the pressure will remain constant. There is no change in the volume of the vessel. Thus, the volume will also remain constant. Since, pressure (P) and volume (V) are constant and R is already a universal constant therefore, the ideal equation can be written as:
$\dfrac{{{n}_{1}}}{{{n}_{2}}}=\dfrac{{{T}_{2}}}{{{T}_{1}}}$
Where,
${{n}_{1}}$ is the initial number of moles,
${{n}_{2}}$ is the final number of moles,
${{T}_{1}}$ is the initial temperature and
${{T}_{2}}$ is the final temperature.
Here, ${{T}_{1}}$ is $300\text{K}$ and ${{T}_{2}}$ is $400\text{K}$ and let’s consider ${{n}_{1}}$ be $1$. So, the fraction of air present in the vessel will be:
$\dfrac{1}{{{n}_{2}}}=\dfrac{{{T}_{2}}}{{{T}_{1}}}$
Then, ${{n}_{2}}=\dfrac{{{T}_{1}}}{{{T}_{2}}}$
Thus, ${{n}_{2}}=\dfrac{300}{400}=\dfrac{3}{4}$
Therefore, the fraction of air present in the vessel is $\dfrac{3}{4}$. Therefore, the gas escaped out will be:
${{n}_{1}}-{{n}_{2}}=1-\dfrac{3}{4}=\dfrac{1}{4}$
Hence, the correct option is B.
Note: Ideal gas law is a good approximation of the behaviour of many gases under many conditions, although it has several limitations. Generally, it is a combination of the empirical Boyle’s law, Charles’ law, Avogadro’s law and Gay-Lussac’s law.
Recently Updated Pages
Master Class 11 Economics: Engaging Questions & Answers for Success

Master Class 11 English: Engaging Questions & Answers for Success

Master Class 11 Social Science: Engaging Questions & Answers for Success

Master Class 11 Biology: Engaging Questions & Answers for Success

Class 11 Question and Answer - Your Ultimate Solutions Guide

Master Class 11 Business Studies: Engaging Questions & Answers for Success

Trending doubts
10 examples of friction in our daily life

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

Difference Between Prokaryotic Cells and Eukaryotic Cells

1 Quintal is equal to a 110 kg b 10 kg c 100kg d 1000 class 11 physics CBSE

Explain zero factorial class 11 maths CBSE

What is a periderm How does periderm formation take class 11 biology CBSE

