
How would you solve for n in $PV=nRT$?
Answer
552.9k+ views
Hint The given equation in the question is the ideal gas equation $PV=nRT$ having n number of moles. When three values in the given equation are given then the fourth value can be calculated because the value of R is constant and it is called a gas constant. By keeping all the given values to one side and keeping the value that is to be calculated on the other side the equation can be solved.
Complete step by step answer:
The given equation in the question is the ideal gas equation $PV=nRT$ having n number of moles. It is solved by combining three laws called Boyle’s law, Charles law, and Avogadro law. When the volume is kept constant, the pressure of the gas is directly proportional to the temperature.
$P\propto T$
When the pressure is kept constant, the volume of the gas is directly proportional to the temperature.
$V\propto T$
When the temperature is kept constant, the pressure of the gas is inversely proportional to the volume.
$P\propto \dfrac{1}{V}$
So, by combining all these and removing the proportionality sign, we got the formula:
$PV=nRT$
In which n is the number of moles of the gas.
When three values in the given equation are given then the fourth value can be calculated because the value of R is constant and it is called a gas constant. By keeping all the given values to one side and keeping the value that is to be calculated on the other side the equation can be solved.
The value of n can be calculated by:
$n=\dfrac{PV}{RT}$
Note: If we want to calculate the pressure of the gas then the equation will become:
$P=\dfrac{nRT}{V}$
If we want to calculate the volume of the gas then the equation will become:
$V=\dfrac{nRT}{P}$
If we want to calculate the temperature of the gas then the equation will become:
$T=\dfrac{PV}{nR}$
Complete step by step answer:
The given equation in the question is the ideal gas equation $PV=nRT$ having n number of moles. It is solved by combining three laws called Boyle’s law, Charles law, and Avogadro law. When the volume is kept constant, the pressure of the gas is directly proportional to the temperature.
$P\propto T$
When the pressure is kept constant, the volume of the gas is directly proportional to the temperature.
$V\propto T$
When the temperature is kept constant, the pressure of the gas is inversely proportional to the volume.
$P\propto \dfrac{1}{V}$
So, by combining all these and removing the proportionality sign, we got the formula:
$PV=nRT$
In which n is the number of moles of the gas.
When three values in the given equation are given then the fourth value can be calculated because the value of R is constant and it is called a gas constant. By keeping all the given values to one side and keeping the value that is to be calculated on the other side the equation can be solved.
The value of n can be calculated by:
$n=\dfrac{PV}{RT}$
Note: If we want to calculate the pressure of the gas then the equation will become:
$P=\dfrac{nRT}{V}$
If we want to calculate the volume of the gas then the equation will become:
$V=\dfrac{nRT}{P}$
If we want to calculate the temperature of the gas then the equation will become:
$T=\dfrac{PV}{nR}$
Recently Updated Pages
Why are manures considered better than fertilizers class 11 biology CBSE

Find the coordinates of the midpoint of the line segment class 11 maths CBSE

Distinguish between static friction limiting friction class 11 physics CBSE

The Chairman of the constituent Assembly was A Jawaharlal class 11 social science CBSE

The first National Commission on Labour NCL submitted class 11 social science CBSE

Number of all subshell of n + l 7 is A 4 B 5 C 6 D class 11 chemistry CBSE

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

What is Environment class 11 chemistry CBSE

Bond order ofO2 O2+ O2 and O22 is in order A O2 langle class 11 chemistry CBSE

How many squares are there in a chess board A 1296 class 11 maths CBSE

Distinguish between verbal and nonverbal communica class 11 english CBSE

The equivalent weight of Mohrs salt FeSO4 NH42SO4 6H2O class 11 chemistry CBSE

