An ideal gas on heating from \[100{{K}}\] to $109{{K}}$ shows an increase by ${{a}}\% $ in its volume at constant ${{P}}$. The value of ${{a}}$ is …………
Answer
Verified
448.2k+ views
Hint: Ideal gas is a type of imaginary gas in which the particles are randomly moving and they do not interact with each other. They are point-like particles. Also, they obey the gas law which is the combination of Boyle’s law, Charles’ law, Gay Lussac law and Avogadro law.
Complete step by step answer:
It is given that the initial temperature, ${{{T}}_1} = 100{{K}}$
Final temperature, ${{{T}}_2} = 109{{K}}$
Percentage increase in volume, ${{V\% = a\% }}$
According to ideal gas equation, ${{PV}} = {{nRT}}$, where ${{P}}$ is the pressure of the gas, ${{V}}$ is the volume of gas, ${{n}}$ is the number of moles of gas, ${{R}}$ is the gas constant and ${{T}}$ is the temperature.
Here, ${{P}},{{n}},{{R}}$ are constants. Since the same ideal gas is heated, there is no change in the number of moles of the gas. Pressure is given as constant.
Thus we can write the ideal gas equation as:
$\dfrac{{{{{V}}_2}}}{{{{{V}}_1}}} = \dfrac{{{{{T}}_2}}}{{{{{T}}_1}}}$, where ${{{V}}_1}$ and ${{{V}}_2}$ are the initial and final volumes of gas respectively.
Substituting the values of initial and final temperatures, we get
$\dfrac{{{{{V}}_2}}}{{{{{V}}_1}}} = \dfrac{{109}}{{100}} \Leftrightarrow {{{V}}_2} = 1.09 \times {{{V}}_1}$
Here, the percentage increase in the volume can be expressed as
$\dfrac{{{{{V}}_2} - {{{V}}_1}}}{{{{{V}}_1}}} \times 100 = {{a}}$
Substituting the value of ${{{V}}_2}$ in the above equation, we get
$\dfrac{{1.09{{{V}}_1} - {{{V}}_1}}}{{{{{V}}_1}}} \times 100 = \dfrac{{9{{{V}}_1}}}{{{{{V}}_1}}} = 9$
Thus the percentage increase in the volume, ${{a}}\% = 9\% $.
So, the value of ${{a}}$ is $9$
Note: Ideal gas equation involves four equations which explain the relationship between volume and pressure at constant number of moles and temperature (Boyle’s law); volume and temperature at constant pressure and number of moles (Charles’ law); pressure and temperature at constant number of moles and volume (Gay Lussac law); number of moles and volume at constant pressure and temperature (Avogadro law).
Complete step by step answer:
It is given that the initial temperature, ${{{T}}_1} = 100{{K}}$
Final temperature, ${{{T}}_2} = 109{{K}}$
Percentage increase in volume, ${{V\% = a\% }}$
According to ideal gas equation, ${{PV}} = {{nRT}}$, where ${{P}}$ is the pressure of the gas, ${{V}}$ is the volume of gas, ${{n}}$ is the number of moles of gas, ${{R}}$ is the gas constant and ${{T}}$ is the temperature.
Here, ${{P}},{{n}},{{R}}$ are constants. Since the same ideal gas is heated, there is no change in the number of moles of the gas. Pressure is given as constant.
Thus we can write the ideal gas equation as:
$\dfrac{{{{{V}}_2}}}{{{{{V}}_1}}} = \dfrac{{{{{T}}_2}}}{{{{{T}}_1}}}$, where ${{{V}}_1}$ and ${{{V}}_2}$ are the initial and final volumes of gas respectively.
Substituting the values of initial and final temperatures, we get
$\dfrac{{{{{V}}_2}}}{{{{{V}}_1}}} = \dfrac{{109}}{{100}} \Leftrightarrow {{{V}}_2} = 1.09 \times {{{V}}_1}$
Here, the percentage increase in the volume can be expressed as
$\dfrac{{{{{V}}_2} - {{{V}}_1}}}{{{{{V}}_1}}} \times 100 = {{a}}$
Substituting the value of ${{{V}}_2}$ in the above equation, we get
$\dfrac{{1.09{{{V}}_1} - {{{V}}_1}}}{{{{{V}}_1}}} \times 100 = \dfrac{{9{{{V}}_1}}}{{{{{V}}_1}}} = 9$
Thus the percentage increase in the volume, ${{a}}\% = 9\% $.
So, the value of ${{a}}$ is $9$
Note: Ideal gas equation involves four equations which explain the relationship between volume and pressure at constant number of moles and temperature (Boyle’s law); volume and temperature at constant pressure and number of moles (Charles’ law); pressure and temperature at constant number of moles and volume (Gay Lussac law); number of moles and volume at constant pressure and temperature (Avogadro law).
Recently Updated Pages
Difference Between Prokaryotic Cells and Eukaryotic Cells
Master Class 12 Business Studies: Engaging Questions & Answers for Success
Master Class 12 English: Engaging Questions & Answers for Success
Master Class 12 Economics: Engaging Questions & Answers for Success
Master Class 12 Chemistry: Engaging Questions & Answers for Success
Master Class 12 Social Science: Engaging Questions & Answers for Success
Trending doubts
One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE
Pigmented layer in the eye is called as a Cornea b class 11 biology CBSE
The lightest gas is A nitrogen B helium C oxygen D class 11 chemistry CBSE
What is spore formation class 11 biology CBSE
In the tincture of iodine which is solute and solv class 11 chemistry CBSE
What are the limitations of Rutherfords model of an class 11 chemistry CBSE