
A gaseous system has a pressure of 580 ${\text{c}}{{\text{m}}^{\text{3}}}$ at a certain pressure. If its pressure is increased by ${\text{0}}{\text{.96 atm}}$, its volume becomes 100${\text{c}}{{\text{m}}^{\text{3}}}$. Determine the pressure of the system.
Answer
570.6k+ views
Hint: This problem is based on Boyle's Law of gas, which states that the pressure of a fixed mass of a gas is inversely proportional to the volume of the gas at constant temperature. We can substitute the given values in the equation to calculate the pressure.
Formula Used:
$\dfrac{{{{\text{P}}_{\text{1}}}}}{{{{\text{P}}_{\text{2}}}}}{\text{ = }}\dfrac{{{{\text{V}}_2}}}{{{{\text{V}}_1}}}$
Complete step by step answer:
As per the question, the initial pressure of the gas, ${{\text{P}}_{\text{1}}}$= P. the initial volume of the gas,${{\text{V}}_{\text{1}}}$ = 580 ${\text{c}}{{\text{m}}^{\text{3}}}$. The pressure increased to${{\text{P}}_2}$= $\left( {{\text{0}}{\text{.96 + P}}} \right)$ and for that the volume changes to ${{\text{V}}_2}$ = 100 ${\text{c}}{{\text{m}}^{\text{3}}}$.
The mathematical expression of the Boyle’s law = \[{{\text{P}}_{\text{1}}} \times {{\text{V}}_{\text{1}}}{\text{ = }}{{\text{P}}_{\text{2}}} \times {{\text{V}}_{\text{2}}}\]
If this equation is rearranged then we get that, $\dfrac{{{{\text{P}}_{\text{1}}}}}{{{{\text{P}}_{\text{2}}}}}{\text{ = }}\dfrac{{{{\text{V}}_2}}}{{{{\text{V}}_1}}}$
Putting the values of all the parameters we get,
$\dfrac{{\text{P}}}{{\left( {{\text{0}}{\text{.96 + P}}} \right)}}{\text{ = }}\dfrac{{100}}{{580}}$
$ \Rightarrow \dfrac{{\text{P}}}{{\left( {{\text{0}}{\text{.96 + P}}} \right)}}{\text{ = }}\dfrac{1}{{5.8}}$
Rearranging:
\[ \Rightarrow {\text{5}}{\text{.8 P = }}\left( {{\text{0}}{\text{.96 + P}}} \right)\]
\[ \Rightarrow {\text{ P = }}\dfrac{{{\text{0}}{\text{.96}}}}{{4.{\text{8}}}} = 0.2\]
Hence, the new pressure of the system initially was equal to $0.2$${\text{atm}}$ respectively and when the volume of the system dropped to 100${\text{c}}{{\text{m}}^{\text{3}}}$, then the pressure of the system becomes ${\text{1}}{\text{.16 atm}}$.
Note:
Most of the gases behave as ideal gases at moderate temperatures and pressures and hence this law is mostly applicable to all types of gases. A beautiful example of the application of Boyle's Law is the breathing system of the Human Body. As we breathe in the air, the volume of the air in the lungs increases and the pressure inside it falls so that that lungs expand to accept the air on the hand as one exhales out the air inside the lungs is released, the volume of air decreases and then the pressure of the lungs is increased.
Formula Used:
$\dfrac{{{{\text{P}}_{\text{1}}}}}{{{{\text{P}}_{\text{2}}}}}{\text{ = }}\dfrac{{{{\text{V}}_2}}}{{{{\text{V}}_1}}}$
Complete step by step answer:
As per the question, the initial pressure of the gas, ${{\text{P}}_{\text{1}}}$= P. the initial volume of the gas,${{\text{V}}_{\text{1}}}$ = 580 ${\text{c}}{{\text{m}}^{\text{3}}}$. The pressure increased to${{\text{P}}_2}$= $\left( {{\text{0}}{\text{.96 + P}}} \right)$ and for that the volume changes to ${{\text{V}}_2}$ = 100 ${\text{c}}{{\text{m}}^{\text{3}}}$.
The mathematical expression of the Boyle’s law = \[{{\text{P}}_{\text{1}}} \times {{\text{V}}_{\text{1}}}{\text{ = }}{{\text{P}}_{\text{2}}} \times {{\text{V}}_{\text{2}}}\]
If this equation is rearranged then we get that, $\dfrac{{{{\text{P}}_{\text{1}}}}}{{{{\text{P}}_{\text{2}}}}}{\text{ = }}\dfrac{{{{\text{V}}_2}}}{{{{\text{V}}_1}}}$
Putting the values of all the parameters we get,
$\dfrac{{\text{P}}}{{\left( {{\text{0}}{\text{.96 + P}}} \right)}}{\text{ = }}\dfrac{{100}}{{580}}$
$ \Rightarrow \dfrac{{\text{P}}}{{\left( {{\text{0}}{\text{.96 + P}}} \right)}}{\text{ = }}\dfrac{1}{{5.8}}$
Rearranging:
\[ \Rightarrow {\text{5}}{\text{.8 P = }}\left( {{\text{0}}{\text{.96 + P}}} \right)\]
\[ \Rightarrow {\text{ P = }}\dfrac{{{\text{0}}{\text{.96}}}}{{4.{\text{8}}}} = 0.2\]
Hence, the new pressure of the system initially was equal to $0.2$${\text{atm}}$ respectively and when the volume of the system dropped to 100${\text{c}}{{\text{m}}^{\text{3}}}$, then the pressure of the system becomes ${\text{1}}{\text{.16 atm}}$.
Note:
Most of the gases behave as ideal gases at moderate temperatures and pressures and hence this law is mostly applicable to all types of gases. A beautiful example of the application of Boyle's Law is the breathing system of the Human Body. As we breathe in the air, the volume of the air in the lungs increases and the pressure inside it falls so that that lungs expand to accept the air on the hand as one exhales out the air inside the lungs is released, the volume of air decreases and then the pressure of the lungs is increased.
Recently Updated Pages
Basicity of sulphurous acid and sulphuric acid are

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

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

Master Class 11 Economics: Engaging Questions & Answers for Success

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

Master Class 11 English: Engaging Questions & Answers for Success

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

Actinoid contraction is more than lanthanoid contraction class 11 chemistry CBSE

The transition element that has lowest enthalpy of class 11 chemistry CBSE

10 examples of friction in our daily life

In Dows process haloarene is converted to phenol with class 11 chemistry CBSE

During the charging of lead storage battery the reaction class 11 chemistry CBSE

