
How do you convert $ 190\,mm\,Hg $ to kilopascals?
Answer
493.5k+ views
Hint: In this question we will apply the unit conversion of the value from $ mm\,\,Hg $ to kilopascals. For this we should know the inter conversion of $ \,1\,atm $ .
We know that
$\Rightarrow 1\,atm = 101.325\,K\,Pa $ .
We know that the common unit of conversion is the atmosphere which is equivalent to $ 760\,mm\,Hg $ and $ 101.3\,K\,Pa $ . We will apply this value and solve the question.
Complete Step By Step Answer:
In the above question, we are given the value of pressure, i.e.
$\Rightarrow P = 190\,mm\,Hg $ .
We know that common units of pressure are related to each other by the following conversion factors:
$\Rightarrow 1\,atm = 101.325\,K\,Pa = 760\,mm\,Hg $
So from the above equation, we can write
$\Rightarrow 101.325\,K\,Pa = 760\,mm\,Hg $
By taking the left hand side term to the right hand side, it gives us
$\Rightarrow \dfrac{{101.325\,K\,Pa}}{{760\,mm\,Hg}} = 1 $ .
Now, we have
$\Rightarrow P = 190\,mm\,Hg $ .
So for $ 190\,mm\,Hg $ , we have the value
$\Rightarrow \dfrac{{101.325\,K\,Pa}}{{760\,mm\,Hg}} \times 190\,mm\,Hg $ .
On solving it gives us value
$\Rightarrow \dfrac{{19251.75}}{{760}} = 25.33\,K\,Pa $
Hence the required value is $ 25\,K\,Pa $ (approx.)
Note:
We should note the above method used is called the dimensional analysis, where a value is multiplied by several conversion factors, before cancelling out the original unit in order to achieve the desired unit. We should also know some of the inter conversion units, such as:
$ 1\,atm = 101325\,Pascal $
$ 1\,atm = 14.6959\,psi $
$ 1\,atm = 1.01325\,bar $ .
We know that
$\Rightarrow 1\,atm = 101.325\,K\,Pa $ .
We know that the common unit of conversion is the atmosphere which is equivalent to $ 760\,mm\,Hg $ and $ 101.3\,K\,Pa $ . We will apply this value and solve the question.
Complete Step By Step Answer:
In the above question, we are given the value of pressure, i.e.
$\Rightarrow P = 190\,mm\,Hg $ .
We know that common units of pressure are related to each other by the following conversion factors:
$\Rightarrow 1\,atm = 101.325\,K\,Pa = 760\,mm\,Hg $
So from the above equation, we can write
$\Rightarrow 101.325\,K\,Pa = 760\,mm\,Hg $
By taking the left hand side term to the right hand side, it gives us
$\Rightarrow \dfrac{{101.325\,K\,Pa}}{{760\,mm\,Hg}} = 1 $ .
Now, we have
$\Rightarrow P = 190\,mm\,Hg $ .
So for $ 190\,mm\,Hg $ , we have the value
$\Rightarrow \dfrac{{101.325\,K\,Pa}}{{760\,mm\,Hg}} \times 190\,mm\,Hg $ .
On solving it gives us value
$\Rightarrow \dfrac{{19251.75}}{{760}} = 25.33\,K\,Pa $
Hence the required value is $ 25\,K\,Pa $ (approx.)
Note:
We should note the above method used is called the dimensional analysis, where a value is multiplied by several conversion factors, before cancelling out the original unit in order to achieve the desired unit. We should also know some of the inter conversion units, such as:
$ 1\,atm = 101325\,Pascal $
$ 1\,atm = 14.6959\,psi $
$ 1\,atm = 1.01325\,bar $ .
Recently Updated Pages
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

Master Class 11 Physics: Engaging Questions & Answers for Success

Master Class 11 Accountancy: Engaging Questions & Answers for Success

Class 11 Question and Answer - Your Ultimate Solutions Guide

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

10 examples of friction in our daily life

