
The volume of mole of a perfect gas at NTP is:
A) $22.4$ $litres$
B) $2.24$ $litres$
C) $100$ $litres$
D) None of these
Answer
135.6k+ views
Hint: In chemistry the mole is an amount unit. One mole of substance is defined as the amount of substance containing the same number of discrete atoms as the number of atoms in a sample of pure $12C$ weighing exactly $12 g$. Experimentally it was determined that one mole of a substance comprises $6.02214179 \times {10^{23}}$ atoms. This number is known as Avagadro’s number.
Complete step by step answer:
We know that the equation for state of an ideal gas is given by,
$PV = nRT$
Where. P is absolute pressure of the gas, V is the volume of gas, T is the temperature of the gas in kelvin. N is the number of moles of the gas and R is the universal gas constant which has a value of $8.314 J/mol-K$. Hence, volume of an ideal gas can be determined by the following equation
$V = \dfrac{{nRT}}{P}$………….(1)
We know that. At N.T.P, temperature is $T$ = $293.15 K \left( {{{20}^o}C} \right)$ and and pressure is $P$ = $1 atm$ = $101.325 kPa$, putting these values along with the value of universal gas constant ($R = 8.314 J/mol-K$) in equation (1), for one mole of ideal gas (n = 1) we get,
$V = \dfrac{{1 \times 8.314 \times 293.15}}{{101.325}}$
$V = 24.054 l$
Hence, at N.T.P the volume of an ideal gas is $24.05$ $litres$. Therefore, we can say that none of the given options are in agreement with our solution.
Hence option D is the correct answer option.
Note: There is a difference between N.T.P and S.T.P and it is mentioned below.
S.T.P stands for standard temperature and pressure. At S.T.P the temperature is $T = {0^o}C$ and pressure is $P = 100 kPa$.
N.T.P stands for normal temperature and pressure. At N.T.P the temperature is $T = {20^o}C$ and pressure is $P =101.325 kPa$.
Complete step by step answer:
We know that the equation for state of an ideal gas is given by,
$PV = nRT$
Where. P is absolute pressure of the gas, V is the volume of gas, T is the temperature of the gas in kelvin. N is the number of moles of the gas and R is the universal gas constant which has a value of $8.314 J/mol-K$. Hence, volume of an ideal gas can be determined by the following equation
$V = \dfrac{{nRT}}{P}$………….(1)
We know that. At N.T.P, temperature is $T$ = $293.15 K \left( {{{20}^o}C} \right)$ and and pressure is $P$ = $1 atm$ = $101.325 kPa$, putting these values along with the value of universal gas constant ($R = 8.314 J/mol-K$) in equation (1), for one mole of ideal gas (n = 1) we get,
$V = \dfrac{{1 \times 8.314 \times 293.15}}{{101.325}}$
$V = 24.054 l$
Hence, at N.T.P the volume of an ideal gas is $24.05$ $litres$. Therefore, we can say that none of the given options are in agreement with our solution.
Hence option D is the correct answer option.
Note: There is a difference between N.T.P and S.T.P and it is mentioned below.
S.T.P stands for standard temperature and pressure. At S.T.P the temperature is $T = {0^o}C$ and pressure is $P = 100 kPa$.
N.T.P stands for normal temperature and pressure. At N.T.P the temperature is $T = {20^o}C$ and pressure is $P =101.325 kPa$.
Recently Updated Pages
JEE Main 2021 July 25 Shift 2 Question Paper with Answer Key

JEE Main 2021 July 25 Shift 1 Question Paper with Answer Key

JEE Main 2021 July 20 Shift 2 Question Paper with Answer Key

JEE Main 2021 July 22 Shift 2 Question Paper with Answer Key

How to find Oxidation Number - Important Concepts for JEE

Half-Life of Order Reactions - Important Concepts and Tips for JEE

Trending doubts
JEE Main 2025 Session 2: Application Form (Out), Exam Dates (Released), Eligibility, & More

JEE Main 2025: Derivation of Equation of Trajectory in Physics

Degree of Dissociation and Its Formula With Solved Example for JEE

A body is falling from a height h After it has fallen class 11 physics JEE_Main

Electric Field Due to Uniformly Charged Ring for JEE Main 2025 - Formula and Derivation

Elastic Collisions in One Dimension - JEE Important Topic

Other Pages
JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

Units and Measurements Class 11 Notes: CBSE Physics Chapter 1

Important Questions for CBSE Class 11 Physics Chapter 1 - Units and Measurement

NCERT Solutions for Class 11 Physics Chapter 2 Motion In A Straight Line

NCERT Solutions for Class 11 Physics Chapter 1 Units and Measurements

Motion In A Plane: Line Class 11 Notes: CBSE Physics Chapter 3
