Molar Specific Heat of gas and relation between them (Mayer's formula) Questions in English

(D) Given that the molar specific heat of a gas is $\frac{5}{2} R$.
We know that for a gas with $f$ degrees of freedom:
$C_V = \frac{fR}{2}$ and $C_P = \left(1 + \frac{f}{2}\right) R$.
Case $1$: If the given specific heat is $C_V$,then $\frac{fR}{2} = \frac{5}{2} R$,which implies $f = 5$. $A$ gas with $f = 5$ is a rigid diatomic gas.
Case $2$: If the given specific heat is $C_P$,then $\left(1 + \frac{f}{2}\right) R = \frac{5}{2} R$. This simplifies to $1 + \frac{f}{2} = 2.5$,so $\frac{f}{2} = 1.5$,which implies $f = 3$. $A$ gas with $f = 3$ is a monoatomic gas.
Since the problem does not specify whether the value is $C_P$ or $C_V$,the gas can be either monoatomic or rigid diatomic.

(C) For a monoatomic gas,the degrees of freedom $f = 3$. The ratio of specific heats is $\gamma_1 = 1 + \frac{2}{f} = 1 + \frac{2}{3} = \frac{5}{3}$.
For a diatomic gas acting as a rigid rotator,the degrees of freedom $f = 5$ ($3$ translational + $2$ rotational). The ratio of specific heats is $\gamma_2 = 1 + \frac{2}{f} = 1 + \frac{2}{5} = \frac{7}{5}$.
Therefore,the ratio $\frac{\gamma_1}{\gamma_2} = \frac{5/3}{7/5} = \frac{5}{3} \times \frac{5}{7} = \frac{25}{21}$.

(D) diatomic gas molecule typically has $3$ translational degrees of freedom and $2$ rotational degrees of freedom.
It is given that the molecule has one additional vibrational mode.
Each vibrational mode contributes $2$ degrees of freedom (one for kinetic energy and one for potential energy).
Therefore, the total number of degrees of freedom $f = 3$ (translational) $+ 2$ (rotational) $+ 2$ (vibrational) $= 7$.
The molar specific heat at constant volume is given by the formula $C_V = \frac{fR}{2}$.
Substituting $f = 7$, we get $C_V = \frac{7R}{2}$.

(B) The ratio of specific heats $\gamma = \frac{C_p}{C_v}$ is defined as $1 + \frac{2}{f}$,where $f$ is the number of degrees of freedom of the gas molecule.
For an ideal gas,the number of degrees of freedom $f$ depends only on the atomicity of the gas (monoatomic,diatomic,or polyatomic) and not on the temperature $T$.
Therefore,$\gamma$ is independent of temperature $T$,which can be expressed as $\gamma \propto T^0$.

(C) For a monoatomic gas,the molar specific heat at constant volume is $(C_v)_{\text{mono}} = \frac{3}{2}R$.
For a rigid diatomic gas,the molar specific heat at constant volume is $(C_v)_{\text{dia}} = \frac{5}{2}R$.
The ratio of the specific heat of these gases at constant volume is given by:
$\frac{(C_v)_{\text{mono}}}{(C_v)_{\text{dia}}} = \frac{\frac{3}{2}R}{\frac{5}{2}R} = \frac{3}{5}$.

(A) For a monoatomic ideal gas:
$C_{v} = \frac{3}{2}R$,$C_{p} = \frac{5}{2}R$.
Thus,$C_{p} - C_{v} = R$,$C_{p} + C_{v} = 4R$,$C_{p}/C_{v} = 5/3 \approx 1.67$,and $C_{p} \cdot C_{v} = 3.75 R^2$.
For a diatomic ideal gas:
$C_{v} = \frac{5}{2}R$,$C_{p} = \frac{7}{2}R$.
Thus,$C_{p} - C_{v} = R$,$C_{p} + C_{v} = 6R$,$C_{p}/C_{v} = 7/5 = 1.4$,and $C_{p} \cdot C_{v} = 8.75 R^2$.
Comparing the values:
$1$. $C_{p} - C_{v} = R$ for both,so $(A)$ is incorrect.
$2$. $C_{p} + C_{v}$ is $6R$ (diatomic) > $4R$ (monoatomic),so $(B)$ is correct.
$3$. $C_{p}/C_{v}$ is $1.4$ (diatomic) < $1.67$ (monoatomic),so $(C)$ is incorrect.
$4$. $C_{p} \cdot C_{v}$ is $8.75 R^2$ (diatomic) > $3.75 R^2$ (monoatomic),so $(D)$ is correct.
Therefore,the correct options are $(B)$ and $(D)$.

$(A)$ The ratio of specific heats is given by $\gamma = \frac{C_P}{C_V} = 1 + \frac{2}{f}$, where $f$ is the degree of freedom.
For a triatomic rigid gas, $f = 6$, so $\gamma = 1 + \frac{2}{6} = \frac{4}{3}$. $(A-III)$
For a diatomic non-rigid gas, $f = 7$, so $\gamma = 1 + \frac{2}{7} = \frac{9}{7}$. $(B-IV)$
For a monoatomic gas, $f = 3$, so $\gamma = 1 + \frac{2}{3} = \frac{5}{3}$. $(C-I)$
For a diatomic rigid gas, $f = 5$, so $\gamma = 1 + \frac{2}{5} = \frac{7}{5}$. $(D-II)$
Thus, the correct match is $A-III, B-IV, C-I, D-II$.

(C) We know that for an ideal gas,the Mayer's relation is given by $C_P - C_V = R$.
Dividing both sides by $C_V$,we get $\frac{C_P}{C_V} - 1 = \frac{R}{C_V}$.
Given that $\gamma = \frac{C_P}{C_V} = \frac{9}{7}$.
Substituting the value of $\gamma$ into the equation,we get $\frac{R}{C_V} = \gamma - 1$.
$\frac{R}{C_V} = \frac{9}{7} - 1 = \frac{9-7}{7} = \frac{2}{7}$.
Calculating the decimal value,$\frac{2}{7} \approx 0.2857$.
Thus,the value is nearly $0.28$.

(C) We know that for an ideal gas,the Mayer's relation is given by: $C_{p} - C_{V} = R$.
Also,the ratio of molar specific heats is defined as: $\gamma = \frac{C_{p}}{C_{V}}$,which implies $C_{V} = \frac{C_{p}}{\gamma}$.
Substituting the value of $C_{V}$ in Mayer's relation:
$C_{p} - \frac{C_{p}}{\gamma} = R$
$C_{p} \left(1 - \frac{1}{\gamma}\right) = R$
$C_{p} \left(\frac{\gamma - 1}{\gamma}\right) = R$
Therefore,$C_{p} = \frac{R \gamma}{\gamma - 1}$.

(A) We know that for an ideal gas,the Mayer's relation is $C_{p} - C_{v} = R$.
Given that $\gamma = \frac{C_{p}}{C_{v}}$,we can write $C_{v} = \frac{C_{p}}{\gamma}$.
Substituting this into the Mayer's relation:
$C_{p} - \frac{C_{p}}{\gamma} = R$
$C_{p} \left(1 - \frac{1}{\gamma}\right) = R$
$C_{p} \left(\frac{\gamma - 1}{\gamma}\right) = R$
Therefore,$C_{p} = \frac{\gamma R}{\gamma - 1}$.

(C) We know that the molar specific heat at constant volume is given by $C_{v} = \frac{n R}{2}$.
From Mayer's relation,$C_{p} - C_{v} = R$,we can write $C_{p} = C_{v} + R$.
Substituting the value of $C_{v}$,we get $C_{p} = \frac{n R}{2} + R = R \left( \frac{n}{2} + 1 \right)$.
Now,the ratio $\gamma = \frac{C_{p}}{C_{v}}$ is calculated as:
$\gamma = \frac{R \left( \frac{n}{2} + 1 \right)}{\frac{n R}{2}} = \frac{\frac{n+2}{2}}{\frac{n}{2}} = \frac{n+2}{n} = 1 + \frac{2}{n}$.

(D) For an ideal gas,the relationship between molar specific heats is given by Mayer's relation:
$C_p - C_v = R$
Dividing both sides of the equation by $C_v$,we get:
$\frac{C_p}{C_v} - 1 = \frac{R}{C_v}$
Since $\gamma = \frac{C_p}{C_v}$,substituting this into the equation:
$\gamma - 1 = \frac{R}{C_v}$
Rearranging the terms to solve for $C_v$:
$C_v = \frac{R}{\gamma - 1}$

(C) For a monoatomic gas,the degrees of freedom $f = 3$. The ratio of specific heats is $\gamma_1 = 1 + \frac{2}{f} = 1 + \frac{2}{3} = \frac{5}{3}$.
For a rigid diatomic gas,the degrees of freedom $f = 5$ ($3$ translational + $2$ rotational). The ratio of specific heats is $\gamma_2 = 1 + \frac{2}{f} = 1 + \frac{2}{5} = \frac{7}{5}$.
Therefore,the ratio $\frac{\gamma_2}{\gamma_1} = \frac{7/5}{5/3} = \frac{7}{5} \times \frac{3}{5} = \frac{21}{25}$.

(D) For an ideal gas,the relationship between molar specific heats at constant pressure $(C_p)$ and constant volume $(C_v)$ is given by Mayer's relation: $C_p - C_v = R$.
We are given the ratio of specific heats as $\gamma = \frac{C_p}{C_v}$,which implies $C_p = \gamma C_v$.
Substituting this into Mayer's relation: $\gamma C_v - C_v = R$.
Factoring out $C_v$: $C_v(\gamma - 1) = R$.
Therefore,$C_v = \frac{R}{\gamma - 1}$.

(D) Given: $C_P = \frac{7}{2} R$,and we know the relation $C_P - C_V = R$.
Therefore,$C_V = C_P - R = \frac{7}{2} R - R = \frac{5}{2} R$.
The adiabatic index $\gamma$ is given by $\gamma = \frac{C_P}{C_V} = \frac{7/2 R}{5/2 R} = \frac{7}{5} = 1.4$.
For a diatomic gas,the degrees of freedom $f = 5$,so $C_V = \frac{f}{2} R = \frac{5}{2} R$ and $C_P = C_V + R = \frac{7}{2} R$.
Thus,the gas consists of diatomic molecules.

(A) The molar specific heat at constant volume,$C_{V}$,is given by the formula $C_{V} = f \times \frac{R}{2}$,where $f$ is the number of degrees of freedom.
Given $f = 6$,we substitute this into the formula:
$C_{V} = 6 \times \frac{R}{2} = 3R$.
Rearranging this equation to solve for $R$,we get:
$R = \frac{C_{V}}{3}$.

(A) For an ideal gas,the relationship between molar specific heats at constant pressure $(C_{P})$ and constant volume $(C_{V})$ is given by Mayer's relation: $C_{P} - C_{V} = R$.
From this,we can write $C_{P} = C_{V} + R$. Therefore,the expression $C_{V} = C_{P} + R$ given in option $(A)$ is incorrect.
Let us verify the other options:
$(B)$ $R = C_{P} - C_{V}$. Since $\gamma = \frac{C_{P}}{C_{V}}$,we have $C_{P} = \gamma C_{V}$. Substituting this,$R = \gamma C_{V} - C_{V} = C_{V}(\gamma - 1)$. This is correct.
$(C)$ $\frac{C_{V}}{C_{P}} = \frac{1}{\gamma}$. Since $\gamma = \frac{C_{P}}{C_{V}}$,this is correct.
$(D)$ $R = C_{P} - C_{V} = C_{P} - \frac{C_{P}}{\gamma} = C_{P}(1 - \frac{1}{\gamma}) = C_{P}(\frac{\gamma - 1}{\gamma})$. This is correct.

(D) The molar heat capacity at constant volume is given by $C_{v} = \frac{n R}{2}$.
The molar heat capacity at constant pressure is given by $C_{p} = C_{v} + R$.
Substituting the value of $C_{v}$,we get $C_{p} = \frac{n R}{2} + R = R \left(1 + \frac{n}{2}\right)$.
The ratio $\gamma = \frac{C_{p}}{C_{v}}$ is calculated as:
$\gamma = \frac{R \left(1 + \frac{n}{2}\right)}{\frac{n R}{2}} = \frac{\frac{2+n}{2}}{\frac{n}{2}} = \frac{2+n}{n} = 1 + \frac{2}{n}$.

(B) Given,$\frac{R}{C_{V}} = 0.4$.
From Mayer's relation,$C_{P} - C_{V} = R$,so $C_{P} = C_{V} + R$.
Substituting $R = 0.4 C_{V}$,we get $C_{P} = C_{V} + 0.4 C_{V} = 1.4 C_{V}$.
The adiabatic index $\gamma$ is defined as $\gamma = \frac{C_{P}}{C_{V}}$.
Therefore,$\gamma = \frac{1.4 C_{V}}{C_{V}} = 1.4$.
For a rigid diatomic gas,the degrees of freedom $f = 5$.
The adiabatic index is $\gamma = 1 + \frac{2}{f} = 1 + \frac{2}{5} = 1 + 0.4 = 1.4$.
Thus,the gas is made up of rigid diatomic molecules.

(C) Given: $\frac{R}{C_{v}} = 0.67$.
We know that the gas constant $R = C_{p} - C_{v}$.
Substituting this into the given equation: $\frac{C_{p} - C_{v}}{C_{v}} = 0.67$.
$\frac{C_{p}}{C_{v}} - 1 = 0.67$.
Since the adiabatic index $\gamma = \frac{C_{p}}{C_{v}}$,we have $\gamma - 1 = 0.67$,which implies $\gamma = 1.67$.
For a monoatomic gas,the degrees of freedom $f = 3$,so $\gamma = 1 + \frac{2}{f} = 1 + \frac{2}{3} = 1 + 0.666... \approx 1.67$.
Therefore,the gas is monoatomic.

(A) We know that for an ideal gas,the molar specific heat at constant pressure is $C_p = \left(1 + \frac{f}{2}\right)R$ and at constant volume is $C_v = \frac{f}{2}R$.
The ratio of specific heats is given by $\gamma = \frac{C_p}{C_v} = 1 + \frac{2}{f}$,which implies $\frac{2}{f} = \gamma - 1$,or $\frac{f}{2} = \frac{1}{\gamma - 1}$.
Substituting this into the expression for $C_p$:
$C_p = \left(1 + \frac{1}{\gamma - 1}\right)R$
$C_p = \left(\frac{\gamma - 1 + 1}{\gamma - 1}\right)R$
$C_p = \frac{\gamma R}{\gamma - 1}$.

(D) According to Mayer's relation for an ideal gas,the difference between molar specific heat at constant pressure $(C_P)$ and constant volume $(C_V)$ is equal to the universal gas constant $(R)$:
$C_P - C_V = R$
We are also given the ratio of molar specific heats as $\gamma = \frac{C_P}{C_V}$,which implies $C_P = \gamma C_V$.
Substituting the expression for $C_P$ into Mayer's relation:
$\gamma C_V - C_V = R$
$C_V(\gamma - 1) = R$
Therefore,$C_V = \frac{R}{\gamma - 1}$.

(C) For a rigid diatomic molecule,the degrees of freedom $f = 5$.
The molar specific heat at constant volume is $C_V = \frac{f}{2} R = \frac{5}{2} R$.
The molar specific heat at constant pressure is $C_P = C_V + R = \frac{5}{2} R + R = \frac{7}{2} R$.
We are given the relation $R = n C_P$.
Substituting the value of $C_P$,we get $R = n (\frac{7}{2} R)$.
Dividing both sides by $R$,we get $1 = n (\frac{7}{2})$.
Therefore,$n = \frac{2}{7} \approx 0.2857$.

(B) The heat energy $Q$ supplied at constant pressure is given by the formula $Q = n C_p \Delta T$.
First,calculate the number of moles $n$ of nitrogen $(N_2)$:
$n = \frac{\text{mass}}{\text{molecular weight}} = \frac{14 \ g}{28 \ g/mol} = 0.5 \ mol$.
Given the change in temperature $\Delta T = 48^{\circ} C$ and the molar heat capacity at constant pressure $C_p = \frac{7}{2} R$.
Substituting these values into the formula:
$Q = 0.5 \times \left(\frac{7}{2} R\right) \times 48$.
$Q = 0.5 \times 3.5 R \times 48$.
$Q = 1.75 R \times 48 = 84 R$.
Therefore,the heat energy supplied is $84 R$.

(B) The fraction of heat energy used to increase the internal energy of the gas is given by the ratio of the change in internal energy $(\Delta U)$ to the total heat supplied $(\Delta Q)$:
$\frac{\Delta U}{\Delta Q} = \frac{n C_v \Delta T}{n C_p \Delta T} = \frac{C_v}{C_p} = \frac{1}{\gamma}$
For an ideal diatomic gas,the adiabatic index $\gamma = \frac{C_p}{C_v} = \frac{7}{5}$.
Therefore,the fraction of energy used to increase internal energy is:
$\frac{\Delta U}{\Delta Q} = \frac{1}{7/5} = \frac{5}{7}$.

(C) For a monoatomic gas at constant pressure,the work done is given by $W = p \Delta V$.
Using the ideal gas equation $pV = nRT$,we have $W = nR \Delta T$.
At constant volume,the heat supplied is equal to the change in internal energy,given by $Q = n C_v \Delta T$.
For a monoatomic gas,the molar heat capacity at constant volume is $C_v = \frac{3}{2} R$.
Substituting this into the heat equation,we get $Q = n \left( \frac{3}{2} R \right) \Delta T$.
Since $W = nR \Delta T$,we can substitute $nR \Delta T$ with $W$.
Therefore,$Q = \frac{3}{2} W$.

(C) For an ideal gas,the heat supplied at constant pressure is given by $\Delta Q = n C_p \Delta T$.
The change in internal energy is given by $\Delta U = n C_v \Delta T$.
The work done at constant pressure is $\Delta W = P \Delta V = n R \Delta T$.
Therefore,the ratio is $\Delta W : \Delta U : \Delta Q = n R \Delta T : n C_v \Delta T : n C_p \Delta T = R : C_v : C_p$.
For a diatomic gas,the degrees of freedom $f = 5$.
Thus,$C_v = \frac{f}{2} R = \frac{5}{2} R$.
Using Mayer's relation,$C_p = C_v + R = \frac{5}{2} R + R = \frac{7}{2} R$.
Substituting these values into the ratio: $R : \frac{5}{2} R : \frac{7}{2} R = 1 : \frac{5}{2} : \frac{7}{2}$.
Multiplying by $2$,we get the ratio $2 : 5 : 7$.

(A) For an ideal gas,the relation between molar specific heats is $C_p - C_v = R$,where $R$ is the universal gas constant.
Since $C_p$ and $C_v$ are given as specific heats (per unit mass),we use the relation $C_p - C_v = \frac{R}{M}$,where $M$ is the molar mass of the gas.
Thus,$R = M(C_p - C_v)$.
The ideal gas equation is $PV = nRT$,where $n = \frac{m}{M}$ is the number of moles.
Substituting $n$ and $R$ into the equation: $PV = \frac{m}{M} \cdot M(C_p - C_v) \cdot T$.
Simplifying,we get $PV = m(C_p - C_v)T$.
Since density $\rho = \frac{m}{V}$,we can rewrite the equation as $P = \frac{m}{V}(C_p - C_v)T$.
Therefore,$\rho = \frac{P}{(C_p - C_v)T}$.

(D) For a monoatomic gas,the molar heat capacities are $C_p = \frac{5}{2}R$ and $C_v = \frac{3}{2}R$.
Work done at constant pressure is $W = nR \Delta T$.
Heat supplied at constant volume is $Q_v = nC_v \Delta T = n \left( \frac{3}{2}R \right) \Delta T$.
Substituting $nR \Delta T = W$ into the expression for $Q_v$,we get $Q_v = \frac{3}{2} W$.

(D) We know that for an ideal gas,the relationship between specific heats at constant pressure $(C_p)$ and constant volume $(C_v)$ is given by Mayer's relation: $C_p - C_v = R$.
Given that the ratio of specific heats is $\gamma = \frac{C_p}{C_v}$,we can write $C_p = \gamma C_v$.
Substituting this into Mayer's relation: $\gamma C_v - C_v = R$.
Factoring out $C_v$: $C_v(\gamma - 1) = R$.
Therefore,$C_v = \frac{R}{\gamma - 1}$.

(A) The ratio of molar specific heats $(\gamma = C_p / C_V)$ is given by the formula $\gamma = 1 + \frac{2}{f}$,where $f$ is the degree of freedom of the gas molecule.
Oxygen $(O_2)$ is a diatomic gas at room temperature.
For a diatomic gas,the number of degrees of freedom is $f = 5$ ($3$ translational and $2$ rotational).
Substituting the value of $f$ into the formula:
$\gamma = 1 + \frac{2}{5} = 1 + 0.4 = 1.4$.

(B) At constant pressure,the heat required is given by $Q_p = n C_p \Delta T$.
Given $n = 2 \ moles$,$\Delta T = 35^{\circ} C - 25^{\circ} C = 10 \ K$,and $Q_p = 310 \ J$.
$310 = 2 \times C_p \times 10 \Rightarrow C_p = \frac{310}{20} = 15.5 \ J \ mol^{-1} K^{-1}$.
Using the relation $C_p - C_V = R$,where $R \approx 8.3 \ J \ mol^{-1} K^{-1}$:
$C_V = C_p - R = 15.5 - 8.3 = 7.2 \ J \ mol^{-1} K^{-1}$.
At constant volume,the heat required is $Q_V = n C_V \Delta T$.
$Q_V = 2 \times 7.2 \times 10 = 144 \ J$.

(C) According to Mayer's relation for an ideal gas:
$C_{p} - C_{V} = R$ $(i)$
where $C_{p}$ is the molar specific heat at constant pressure and $C_{V}$ is the molar specific heat at constant volume.
By definition,the ratio of specific heats is:
$\gamma = \frac{C_{p}}{C_{V}}$
This implies $C_{p} = \gamma C_{V}$ (ii)
Substituting equation (ii) into equation $(i)$:
$\gamma C_{V} - C_{V} = R$
$C_{V}(\gamma - 1) = R$
Therefore,$C_{V} = \frac{R}{\gamma - 1}$

(A) We are given the ratio $\frac{R}{C_v} = 0.67$.
We know that the Mayer's relation is $C_p - C_v = R$,which can be written as $\frac{C_p}{C_v} - 1 = \frac{R}{C_v}$.
Let $\gamma = \frac{C_p}{C_v}$ be the adiabatic index.
Then,$\gamma - 1 = 0.67$,which implies $\gamma = 1.67$.
For a monoatomic gas,$\gamma = \frac{5}{3} \approx 1.67$.
For a diatomic gas,$\gamma = \frac{7}{5} = 1.4$.
For a polyatomic gas,$\gamma < 1.4$.
Since $\gamma = 1.67$,the gas is monoatomic.

(B) For a diatomic gas,the degrees of freedom $f_1 = 5$.
The molar specific heat capacity at constant volume is $C_{v_1} = \frac{f_1 R}{2} = \frac{5 R}{2}$.
The molar specific heat capacity at constant pressure is $C_{p_1} = C_{v_1} + R = \frac{5 R}{2} + R = \frac{7 R}{2} = C$.
For a monoatomic gas,the degrees of freedom $f_2 = 3$.
The molar specific heat capacity at constant volume is $C_{v_2} = \frac{f_2 R}{2} = \frac{3 R}{2}$.
Now,taking the ratio $\frac{C}{C_{v_2}} = \frac{\frac{7 R}{2}}{\frac{3 R}{2}} = \frac{7}{3}$.
Therefore,$C_{v_2} = \frac{3 C}{7}$.

(A) The ratio of specific heats is given by $\gamma = \frac{C_P}{C_V} = 1 + \frac{2}{f}$.
For Monoatomic gas: $f=3$,so $\gamma = 1 + \frac{2}{3} = \frac{5}{3}$. Thus,$A-II$.
For Diatomic (rigid) gas: $f=5$,so $\gamma = 1 + \frac{2}{5} = \frac{7}{5}$. Thus,$B-III$.
For Diatomic (non-rigid) gas: $f=7$,so $\gamma = 1 + \frac{2}{7} = \frac{9}{7}$. Thus,$C-IV$.
For Polyatomic gas: The general formula for $\gamma$ is $\frac{4+f}{3+f}$. Thus,$D-I$.
Therefore,the correct matching is $A-II, B-III, C-IV, D-I$.

(B) For a monatomic gas,the degree of freedom $f = 3$.
$\therefore (C_{p})_1 = (1 + \frac{f}{2}) R = (1 + \frac{3}{2}) R = \frac{5 R}{2}$.
For a diatomic gas,the degree of freedom $f = 5$.
$\therefore (C_{p})_2 = (1 + \frac{f}{2}) R = (1 + \frac{5}{2}) R = \frac{7 R}{2}$.
Therefore,the ratio is $\frac{(C_{p})_1}{(C_{p})_2} = \frac{5 R / 2}{7 R / 2} = 5 : 7$.

(C) For a monatomic gas,the molar specific heat capacity at constant volume is $C_V = \frac{3}{2} R$.
For a monatomic gas,the molar specific heat capacity at constant pressure is $C_P = \frac{5}{2} R$.
According to the problem,$C_V = \frac{x}{100} \times C_P$.
Substituting the values,we get $\frac{3}{2} R = \frac{x}{100} \times \frac{5}{2} R$.
Canceling $\frac{1}{2} R$ from both sides,we get $3 = \frac{x}{100} \times 5$.
Solving for $x$,we get $x = \frac{3 \times 100}{5} = 60$.

(C) For a non-rigid diatomic molecule,the degrees of freedom $(f)$ are calculated as follows:
Translational degrees of freedom $= 3$
Rotational degrees of freedom $= 2$
Vibrational degrees of freedom $= 2$ (one for kinetic energy and one for potential energy).
Total degrees of freedom $(f) = 3 + 2 + 2 = 7$.
The ratio of specific heats is given by $\gamma = \frac{C_p}{C_v} = 1 + \frac{2}{f}$.
Substituting $f = 7$,we get $\gamma = 1 + \frac{2}{7} = \frac{9}{7}$.
Therefore,$\frac{C_p}{C_v} = \frac{9}{7}$,which implies $7 C_p = 9 C_v$.
Squaring both sides,we get $49 C_p^2 = 81 C_v^2$.

(C) The heat supplied at constant pressure is $dQ_p = n C_p \Delta T$.
The change in internal energy is $dU = n C_v \Delta T$.
The fraction of heat energy used to increase the internal energy is given by the ratio $\frac{dU}{dQ_p} = \frac{n C_v \Delta T}{n C_p \Delta T} = \frac{C_v}{C_p} = \frac{1}{\gamma}$.
For a diatomic gas,the degrees of freedom $f = 5$.
The molar heat capacity at constant volume is $C_v = \frac{f}{2} R = \frac{5}{2} R$.
The molar heat capacity at constant pressure is $C_p = C_v + R = \frac{5}{2} R + R = \frac{7}{2} R$.
Therefore,the required fraction is $\frac{C_v}{C_p} = \frac{\frac{5}{2} R}{\frac{7}{2} R} = \frac{5}{7}$.

(C) From Mayer's formula,we know that $C_p - C_V = R$,which implies $C_p = C_V + R$.
At constant volume,the heat capacity $C_V$ is defined as the rate of change of internal energy with respect to temperature:
$C_V = \frac{d U}{d T}$.
Substituting this expression for $C_V$ into Mayer's formula,we get:
$C_p = \frac{d U}{d T} + R$.

(C) For a triatomic non-linear gas,the degree of freedom is $f = 6$.
The molar heat capacity at constant volume is $C_V = \frac{f R}{2} = \frac{6 R}{2} = 3 R$.
The molar heat capacity at constant pressure is $C_p = C_V + R = 3 R + R = 4 R$.
The ratio of adiabatic elasticity to isothermal elasticity is equal to the adiabatic index $\gamma$.
$\gamma = \frac{C_p}{C_V} = \frac{4 R}{3 R} = \frac{4}{3}$.
Thus,the ratio is $4: 3$.

(B) We know that for an ideal gas,the relationship between the molar heat capacity at constant volume $(C_v)$,the universal gas constant $(R)$,and the adiabatic index $(\gamma)$ is given by Mayer's relation: $C_p - C_v = R$.
Dividing by $C_v$,we get $\frac{C_p}{C_v} - 1 = \frac{R}{C_v}$.
Since $\gamma = \frac{C_p}{C_v}$,we have $\gamma - 1 = \frac{R}{C_v}$.
Given that $\frac{R}{C_v} = 0.4$,we substitute this into the equation: $\gamma - 1 = 0.4$,which gives $\gamma = 1.4$.
$A$ value of $\gamma = 1.4$ corresponds to a diatomic gas.

(D) Given,specific heat capacity at constant pressure,$C_p = 620 \ J \ kg^{-1} \ K^{-1}$ and specific heat capacity at constant volume,$C_v = 420 \ J \ kg^{-1} \ K^{-1}$.
The molar mass $M$ of the gas is related to the gas constant $R$ by the relation: $M(C_p - C_v) = R$.
Substituting the values: $M(620 - 420) = 8.314 \ J \ mol^{-1} \ K^{-1}$.
$M(200) = 8.314 \implies M = \frac{8.314}{200} = 0.04157 \ kg \ mol^{-1}$.
At $STP$,pressure $P = 1.013 \times 10^5 \ Pa$ and temperature $T = 273.15 \ K$.
Using the ideal gas equation $PV = nRT = \frac{m}{M}RT$,we get the density $\rho = \frac{m}{V} = \frac{PM}{RT}$.
$\rho = \frac{(1.013 \times 10^5) \times 0.04157}{8.314 \times 273.15} \approx 1.855 \ kg \ m^{-3}$.
Rounding to the nearest value,we get $\rho \approx 1.86 \ kg \ m^{-3}$.
Therefore,option $(D)$ is correct.

(D) Helium is a monoatomic gas. For an ideal gas,the relationship between molar specific heat at constant pressure $(C_p)$ and molar specific heat at constant volume $(C_V)$ is given by Mayer's relation: $C_p - C_V = R$.
Given: $C_V = 12.6 \ J \ mol^{-1} \ K^{-1}$ and $R = 8.314 \ J \ mol^{-1} \ K^{-1}$.
Substituting these values into the equation: $C_p = C_V + R$.
$C_p = 12.6 + 8.314 = 20.914 \ J \ mol^{-1} \ K^{-1}$.
Rounding this value to the nearest integer,we get $C_p \approx 21 \ J \ mol^{-1} \ K^{-1}$.

(C) For a monatomic gas,the degrees of freedom $f = 3$.
The molar heat capacity at constant volume is $C_v = \frac{f}{2}R = \frac{3}{2}R$.
The molar heat capacity at constant pressure is $C_p = C_v + R = \frac{5}{2}R$.
The heat absorbed at constant pressure is $Q = n C_p \Delta T = 40 \ J$.
The change in internal energy is $\Delta U = n C_v \Delta T$.
Taking the ratio,$\frac{\Delta U}{Q} = \frac{n C_v \Delta T}{n C_p \Delta T} = \frac{C_v}{C_p} = \frac{3/2 R}{5/2 R} = \frac{3}{5}$.
Therefore,$\Delta U = \frac{3}{5} \times Q = \frac{3}{5} \times 40 \ J = 24 \ J$.

(A) At $STP$, the molar volume of an ideal gas is $22.4 \text{ litres/mol}$.
Number of moles of helium gas, $n = \frac{67.2 \text{ L}}{22.4 \text{ L/mol}} = 3 \text{ mol}$.
Helium is a monoatomic gas, so its molar heat capacity at constant volume is $C_v = \frac{3}{2} R$.
The heat required to raise the temperature by $dT = 20^{\circ} C$ (or $20 \text{ K}$) at constant volume is given by:
$dQ = n C_v dT$
$dQ = 3 \times \left( \frac{3}{2} \times 8.314 \text{ J/mol K} \right) \times 20 \text{ K}$
$dQ = 3 \times 1.5 \times 8.314 \times 20$
$dQ = 9 \times 83.14 = 748.26 \text{ J} \approx 748 \text{ J}$.
Given the options, the closest value is $748 \text{ J}$ (Note: Using $R = 8.31 \text{ J/mol K}$ gives $747.9 \text{ J}$, and using $R = 8.3 \text{ J/mol K}$ gives $747 \text{ J}$. The provided option $784 \text{ J}$ appears to be a calculation approximation or typo in the source, but we select the closest logical result).