$\gamma_A$ is the specific heat ratio of monoatomic gas A having 3 translational degrees of freedom. $\gamma_B$ is the specific heat ratio of polyatomic gas B having 3 translational, 3 rotational degrees of freedom and 1 vibrational mode. If \[ \frac{\gamma_A}{\gamma_B} = \left( 1 + \frac{1}{n} \right) \] then the value of \( n \) is ___.
$\gamma_A$ is the specific heat ratio of monoatomic gas A having 3 translational degrees of freedom. $\gamma_B$ is the specific heat ratio of polyatomic gas B having 3 translational, 3 rotational degrees of freedom and 1 vibrational mode. If \[ \frac{\gamma_A}{\gamma_B} = \left( 1 + \frac{1}{n} \right) \] then the value of \( n \) is ___.
Show Hint
Correct Answer: 3
Approach Solution - 1
Approach Solution -2
The problem asks to find the value of \( n \) from the given relation \( \frac{\gamma_A}{\gamma_B} = \left( 1 + \frac{1}{n} \right) \), where \( \gamma_A \) and \( \gamma_B \) are the specific heat ratios for a monoatomic gas A and a polyatomic gas B, respectively, with their degrees of freedom specified.
Concept Used:
The specific heat ratio (or adiabatic index), \( \gamma \), of a gas is related to its total degrees of freedom, \( f \), by the formula:
\[ \gamma = 1 + \frac{2}{f} \]
The total degrees of freedom (\( f \)) is the sum of translational, rotational, and vibrational degrees of freedom.
- Each translational degree of freedom contributes \( \frac{1}{2}kT \) to the internal energy.
- Each rotational degree of freedom contributes \( \frac{1}{2}kT \) to the internal energy.
- Each vibrational mode contributes \( kT \) to the internal energy ( \( \frac{1}{2}kT \) for kinetic and \( \frac{1}{2}kT \) for potential energy), which corresponds to 2 degrees of freedom.
Step-by-Step Solution:
Step 1: Calculate the degrees of freedom for the monoatomic gas A (\( f_A \)).
A monoatomic gas has only translational motion. It is given that there are 3 translational degrees of freedom.
\[ f_A = f_{\text{trans}} = 3 \]
Step 2: Calculate the specific heat ratio for gas A (\( \gamma_A \)).
Using the formula for \( \gamma \):
\[ \gamma_A = 1 + \frac{2}{f_A} = 1 + \frac{2}{3} = \frac{3+2}{3} = \frac{5}{3} \]
Step 3: Calculate the degrees of freedom for the polyatomic gas B (\( f_B \)).
Gas B has 3 translational, 3 rotational, and 1 vibrational mode. As each vibrational mode corresponds to 2 degrees of freedom:
\[ f_B = f_{\text{trans}} + f_{\text{rot}} + f_{\text{vib}} \] \[ f_B = 3 + 3 + (1 \times 2) = 8 \]
Step 4: Calculate the specific heat ratio for gas B (\( \gamma_B \)).
Using the formula for \( \gamma \):
\[ \gamma_B = 1 + \frac{2}{f_B} = 1 + \frac{2}{8} = 1 + \frac{1}{4} = \frac{4+1}{4} = \frac{5}{4} \]
Step 5: Use the given relation to find the value of \( n \).
The relation is:
\[ \frac{\gamma_A}{\gamma_B} = 1 + \frac{1}{n} \]
Substitute the calculated values of \( \gamma_A \) and \( \gamma_B \):
\[ \frac{5/3}{5/4} = 1 + \frac{1}{n} \] \[ \frac{5}{3} \times \frac{4}{5} = 1 + \frac{1}{n} \] \[ \frac{4}{3} = 1 + \frac{1}{n} \]
Step 6: Solve for \( n \).
\[ \frac{1}{n} = \frac{4}{3} - 1 \] \[ \frac{1}{n} = \frac{4 - 3}{3} = \frac{1}{3} \] \[ n = 3 \]
The value of \( n \) is 3.
Top JEE Main Physics Questions
- In a hydrogen like atom, when an electron jumps from the $M$ - shell to the $L$ - shell, the wavelength of emitted radiation is $\lambda$. If an electron jumps from $N$-shell to the $L$-shell, the wavelength of emitted radiation will be :
- Two masses $m$ and $\frac{m}{2}$ are connected at the two ends of a massless rigid rod of length $l$. The rod is suspended by a thin wire of torsional constant $k$ at the centre of mass of the rod-mass system(see figure). Because of torsional constant $k$, the restoring torque is $\tau =k\theta$ for angular displacement $\theta$. If the rod is rota ted by $\theta_0$ and released, the tension in it when it passes through its mean position will be:
A black body is at a temperature of 2880 K. The energy of radiation emitted by this body with wavelength between 499 nm and 500 nm is U1, between 999 nm and 1000 nm is U2 and between 1499 nm and 1500 nm is U3. The Wien's constant, b = 2.88×106 nm-K. Then,
- $I _{ CM }$ is the moment of inertia of a circular disc about an axis (CM) passing through its center and perpendicular to the plane of disc $I_{A B}$ is it's moment of inertia about an axis $AB$ perpendicular to plane and parallel to axis $CM$ at a distance $\frac{2}{3} R$ from center Where $R$ is the radius of the dise The ratio of $I _{ AB }$ and $I _{ CM }$ is $x: 9$ The value of $x$ is______
- A nucleus disintegrates into two smaller parts, which have their velocities in the ratio $3: 2$ The ratio of their nuclear sizes will be $\left(\frac{x}{3}\right)^{\frac{1}{3}}$ The value of ' $x$ ' is:-
Top JEE Main Thermodynamics Questions
- An electric appliance supplies 6000 J/min heat to the system. If the system delivers a power of 90 W. How long it would take to increase the internal energy by $2.5 \times 10^3$ J ?
- The difference of temperature in a material can convert heat energy into electrical energy. To harvest the heat energy, the material should have:
- Given below are two statements. One is labelled as Assertion (A) and the other is labelled as Reason (R).
Assertion (A): With the increase in the pressure of an ideal gas, the volume falls off more rapidly in an isothermal process in comparison to the adiabatic process.
Reason (R): In an isothermal process, \( PV = \text{constant} \), while in an adiabatic process \( PV^\gamma = \text{constant} \). Here, \( \gamma \) is the ratio of specific heats, \( P \) is the pressure and \( V \) is the volume of the ideal gas.
In the light of the above statements, choose the correct answer from the options given below: - A poly-atomic molecule (C\(_3\)R, \(C_v = 4R\), where \(R\) is gas constant) goes from phase space point A (\(P_A = 10^4 \, \text{Pa}, V_A = 4 \times 10^{-3} \, \text{m}^3\)) to point B (\(P_B = 5 \times 10^4 \, \text{Pa}, V_B = 6 \times 10^{-7} \, \text{m}^3\)) to point C (\(P_C = 10^4 \, \text{Pa}, V_C = 8 \times 10^{-3} \, \text{m}^3\)). A to B is an adiabatic path and B to C is an isothermal path. The net heat absorbed per unit mole by the system is:
- Water falls from a height of 200 m into a pool. Calculate the rise in temperature of the water assuming no heat dissipation from the water in the pool. (Take $ g = 10 \, \text{m/s}^2 $, specific heat of water = 4200 J/(kg K))
Top JEE Main Questions
- In a hydrogen like atom, when an electron jumps from the $M$ - shell to the $L$ - shell, the wavelength of emitted radiation is $\lambda$. If an electron jumps from $N$-shell to the $L$-shell, the wavelength of emitted radiation will be :
- Two masses $m$ and $\frac{m}{2}$ are connected at the two ends of a massless rigid rod of length $l$. The rod is suspended by a thin wire of torsional constant $k$ at the centre of mass of the rod-mass system(see figure). Because of torsional constant $k$, the restoring torque is $\tau =k\theta$ for angular displacement $\theta$. If the rod is rota ted by $\theta_0$ and released, the tension in it when it passes through its mean position will be:
What will be the equilibrium constant of the given reaction carried out in a \(5 \,L\) vessel and having equilibrium amounts of \(A_2\) and \(A\) as \(0.5\) mole and \(2 \times 10^{-6}\) mole respectively?
The reaction : \(A_2 \rightleftharpoons 2A\)- The value of is
- The number of terms of an is even. The sum of the odd terms is and of the even terms is . The last term exceeds the first by . Then the number of terms in the series is ______.