How many degrees of freedom are associated with 2 grams of He at NTP?
A) 3
B) $3 \cdot 01 \times {10^{23}}$
C) $9 \cdot 03 \times {10^{23}}$
D) 6
Answer
Verified
123k+ views
Hint: Firstly we calculate the number of moles in 2 gram of helium .
After that we calculate number of molecules of helium in given number of moles
Now we already know that there are 3 degrees of freedom corresponding to 1 molecule of a monatomic gas.
Finally to calculate the total number of degrees of freedom in a monatomic gas we multiply the number of molecules and degree of freedom of 1 monatomic gas.
Complete step by step process:
According to the question we have 2 gm of helium.
Number of moles =given mass of substance divided by molar mass.
$\therefore $we already know the molar mass of He is 4 amu
So, moles of He=$\dfrac{2}{4} = \dfrac{1}{2}$
Now to calculate moles into molecules we multiply moles with the Avogadro's number
Mathematically, $N = m \times {A_0}$ where N=number of molecules
M=number of moles
${A_o}$=Avogadro's number
So ,$N = 6 \cdot 02 \times {10^{^{23}}} \times \dfrac{1}{2}$
$N = 3 \cdot 01 \times {10^{23}}$
Hence total number of molecules in $\dfrac{1}{2}$moles of He is $3 \cdot 01 \times {10^{23}}$
Now total degree of freedom is equal to molecules multiply by degree of freedom of 1 monatomic gas
$\therefore $total degree of freedom =$3 \times 3 \cdot 01 \times {10^{23}}$
Total degree of freedom =$9 \cdot 03 \times {10^{23}}$.
Hence, option (C) is the best option.
Note: Degree of freedom, often abbreviated as df, is a concept that may be thought of as that part of the sample size n not otherwise allocated. Df is related to the sample number, usually to the number of observations for continuous data methods and to the number of categories for categorical data methods.
After that we calculate number of molecules of helium in given number of moles
Now we already know that there are 3 degrees of freedom corresponding to 1 molecule of a monatomic gas.
Finally to calculate the total number of degrees of freedom in a monatomic gas we multiply the number of molecules and degree of freedom of 1 monatomic gas.
Complete step by step process:
According to the question we have 2 gm of helium.
Number of moles =given mass of substance divided by molar mass.
$\therefore $we already know the molar mass of He is 4 amu
So, moles of He=$\dfrac{2}{4} = \dfrac{1}{2}$
Now to calculate moles into molecules we multiply moles with the Avogadro's number
Mathematically, $N = m \times {A_0}$ where N=number of molecules
M=number of moles
${A_o}$=Avogadro's number
So ,$N = 6 \cdot 02 \times {10^{^{23}}} \times \dfrac{1}{2}$
$N = 3 \cdot 01 \times {10^{23}}$
Hence total number of molecules in $\dfrac{1}{2}$moles of He is $3 \cdot 01 \times {10^{23}}$
Now total degree of freedom is equal to molecules multiply by degree of freedom of 1 monatomic gas
$\therefore $total degree of freedom =$3 \times 3 \cdot 01 \times {10^{23}}$
Total degree of freedom =$9 \cdot 03 \times {10^{23}}$.
Hence, option (C) is the best option.
Note: Degree of freedom, often abbreviated as df, is a concept that may be thought of as that part of the sample size n not otherwise allocated. Df is related to the sample number, usually to the number of observations for continuous data methods and to the number of categories for categorical data methods.
Recently Updated Pages
The ratio of the diameters of two metallic rods of class 11 physics JEE_Main
What is the difference between Conduction and conv class 11 physics JEE_Main
Mark the correct statements about the friction between class 11 physics JEE_Main
Find the acceleration of the wedge towards the right class 11 physics JEE_Main
A standing wave is formed by the superposition of two class 11 physics JEE_Main
Derive an expression for work done by the gas in an class 11 physics JEE_Main
Trending doubts
JEE Mains 2025: Check Important Dates, Syllabus, Exam Pattern, Fee and Updates
JEE Main Login 2045: Step-by-Step Instructions and Details
JEE Main Chemistry Question Paper with Answer Keys and Solutions
JEE Main Exam Marking Scheme: Detailed Breakdown of Marks and Negative Marking
JEE Main 2023 January 24 Shift 2 Question Paper with Answer Keys & Solutions
JEE Main Chemistry Exam Pattern 2025
Other Pages
JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs
NCERT Solutions for Class 11 Physics Chapter 1 Units and Measurements
NCERT Solutions for Class 11 Physics Chapter 9 Mechanical Properties of Fluids
Units and Measurements Class 11 Notes: CBSE Physics Chapter 1
JEE Advanced 2025: Dates, Registration, Syllabus, Eligibility Criteria and More
NCERT Solutions for Class 11 Physics Chapter 2 Motion In A Straight Line