
How much should the pressure be increased in order to decrease the volume of the gas by 5% at a constant temperature?
Answer
582.3k+ views
Hint: When a gas expands its pressure decreases and its volume increases. This means that the pressure of the gas is inversely proportional to the volume of the gas. ${{P}_{1}}{{V}_{1}}={{P}_{2}}{{V}_{2}}$ can be used to solve this problem.
Complete step by step answer:
According to the question the pressure is increasing and volume is decreasing at a constant temperature, this corresponds to Boyle's law. It states that the pressure and volume of the gas are inversely proportional to each other.
It can be formulated as:
$P\propto \dfrac{1}{V}$
$PV=K$
Where K is constant.
This relation can be converted into:
${{P}_{1}}{{V}_{1}}={{P}_{2}}{{V}_{2}}$
So according to the question ${{V}_{1}}$ is V and ${{V}_{2}}$ is 5% less than the original volume at a constant temperature.
For ${{V}_{2}}$ can be written as:
${{V}_{2}}=(100-5)%{{V}_{1}}$
${{V}_{2}}=95%\text{ }{{V}_{1}}$
${{V}_{2}}=0.95\text{ }{{V}_{1}}$
And we have to find the increase in pressure:
Putting the values of both volumes in the equation, we get:
${{P}_{1}}{{V}_{1}}={{P}_{2}}\text{ 0}\text{.95 }{{V}_{1}}$
From this we get the relation of pressure:
${{P}_{1}}={{P}_{2}}\text{ x 0}\text{.95}$
So ${{P}_{1}}$ is equal to 0.95 times the ${{P}_{2}}$
This can also be written as:
${{P}_{2}}=\dfrac{{{P}_{1}}}{0.95}=1.0526{{P}_{1}}$
Or we can say that ${{P}_{2}}$ is equal to 1.0526 times the ${{P}_{1}}$
We know that change in pressure is equal to the difference in the final pressure and the initial pressure.
$={{P}_{2}}-{{P}_{1}}$
We know the value of ${{P}_{2}}$, so putting the value of ${{P}_{2}}$ in this equation we get:
$=1.526{{P}_{1}}-{{P}_{1}}=0.0526{{P}_{1}}$
On converting this into percentage, we get
$0.0526{{P}_{1}}\text{ x 100 = 5}\text{.26 }\!\!%\!\!\text{ }{{\text{P}}_{1}}$
So, 5.26% of pressure must be increased in order to decrease the volume of the gas by 5%.
Note: The formula used in this equation only if the condition is specified at a constant temperature. Because this formula is based on Boyle's law.
Complete step by step answer:
According to the question the pressure is increasing and volume is decreasing at a constant temperature, this corresponds to Boyle's law. It states that the pressure and volume of the gas are inversely proportional to each other.
It can be formulated as:
$P\propto \dfrac{1}{V}$
$PV=K$
Where K is constant.
This relation can be converted into:
${{P}_{1}}{{V}_{1}}={{P}_{2}}{{V}_{2}}$
So according to the question ${{V}_{1}}$ is V and ${{V}_{2}}$ is 5% less than the original volume at a constant temperature.
For ${{V}_{2}}$ can be written as:
${{V}_{2}}=(100-5)%{{V}_{1}}$
${{V}_{2}}=95%\text{ }{{V}_{1}}$
${{V}_{2}}=0.95\text{ }{{V}_{1}}$
And we have to find the increase in pressure:
Putting the values of both volumes in the equation, we get:
${{P}_{1}}{{V}_{1}}={{P}_{2}}\text{ 0}\text{.95 }{{V}_{1}}$
From this we get the relation of pressure:
${{P}_{1}}={{P}_{2}}\text{ x 0}\text{.95}$
So ${{P}_{1}}$ is equal to 0.95 times the ${{P}_{2}}$
This can also be written as:
${{P}_{2}}=\dfrac{{{P}_{1}}}{0.95}=1.0526{{P}_{1}}$
Or we can say that ${{P}_{2}}$ is equal to 1.0526 times the ${{P}_{1}}$
We know that change in pressure is equal to the difference in the final pressure and the initial pressure.
$={{P}_{2}}-{{P}_{1}}$
We know the value of ${{P}_{2}}$, so putting the value of ${{P}_{2}}$ in this equation we get:
$=1.526{{P}_{1}}-{{P}_{1}}=0.0526{{P}_{1}}$
On converting this into percentage, we get
$0.0526{{P}_{1}}\text{ x 100 = 5}\text{.26 }\!\!%\!\!\text{ }{{\text{P}}_{1}}$
So, 5.26% of pressure must be increased in order to decrease the volume of the gas by 5%.
Note: The formula used in this equation only if the condition is specified at a constant temperature. Because this formula is based on Boyle's law.
Recently Updated Pages
Master Class 11 Computer Science: Engaging Questions & Answers for Success

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

Master Class 11 Economics: Engaging Questions & Answers for Success

Master Class 11 English: Engaging Questions & Answers for Success

Master Class 11 Maths: Engaging Questions & Answers for Success

Master Class 11 Biology: Engaging Questions & Answers for Success

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

There are 720 permutations of the digits 1 2 3 4 5 class 11 maths CBSE

Discuss the various forms of bacteria class 11 biology CBSE

Draw a diagram of a plant cell and label at least eight class 11 biology CBSE

State the laws of reflection of light

Explain zero factorial class 11 maths CBSE

