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

(B) For an ideal gas,the adiabatic elasticity $E_{\phi}$ is given by $E_{\phi} = \gamma P$,where $\gamma$ is the adiabatic index and $P$ is the pressure.
The isothermal elasticity $E_{\theta}$ is given by $E_{\theta} = P$.
The ratio of adiabatic elasticity to isothermal elasticity is $\frac{E_{\phi}}{E_{\theta}} = \frac{\gamma P}{P} = \gamma$.
Since $\gamma = \frac{C_p}{C_v}$,the ratio is $\frac{C_p}{C_v}$.

(B) The adiabatic elasticity of a gas is given by $E_{\text{adiabatic}} = \gamma P$,where $\gamma$ is the adiabatic index and $P$ is the pressure.
The isothermal elasticity of a gas is given by $E_{\text{isothermal}} = P$.
The ratio of adiabatic to isothermal elasticity is $\frac{E_{\text{adiabatic}}}{E_{\text{isothermal}}} = \frac{\gamma P}{P} = \gamma$.
For a triatomic gas (non-linear),the degrees of freedom $f = 6$. The adiabatic index $\gamma = 1 + \frac{2}{f} = 1 + \frac{2}{6} = 1 + \frac{1}{3} = \frac{4}{3}$.
Therefore,the ratio is $\frac{4}{3}$.

(A) The ratio of specific heats is defined as $\gamma = \frac{C_P}{C_V}$.
According to the kinetic theory of gases,the molar specific heat at constant volume is $C_V = \frac{f}{2}R$.
The molar specific heat at constant pressure is $C_P = C_V + R = \frac{f}{2}R + R = R(1 + \frac{f}{2})$.
Therefore,the ratio $\gamma = \frac{C_P}{C_V} = \frac{R(1 + f/2)}{(f/2)R} = \frac{1 + f/2}{f/2} = \frac{2}{f} + 1 = 1 + \frac{2}{f}$.

(D) For a diatomic gas molecule considering translational,rotational,and vibrational degrees of freedom:
Translational degrees of freedom $(f_t)$ = $3$.
Rotational degrees of freedom $(f_r)$ = $2$.
Vibrational degrees of freedom $(f_v)$ = $1$ (which contributes $2$ to the total degrees of freedom as it includes both kinetic and potential energy).
Total degrees of freedom $(f)$ = $f_t + f_r + 2f_v = 3 + 2 + 2(1) = 7$.
However,if the question implies $1$ vibrational mode as a single degree of freedom contribution (often simplified in specific contexts),then $f = 3 + 2 + 1 = 6$.
Using the formula $\gamma = \frac{C_P}{C_V} = 1 + \frac{2}{f}$:
For $f = 6$,$\gamma = 1 + \frac{2}{6} = 1 + \frac{1}{3} = \frac{4}{3} \approx 1.33$.

(B) Neon $(Ne)$ is a monoatomic gas.
For a monoatomic gas,the degrees of freedom $(f)$ is $3$.
The molar specific heat at constant volume is given by the formula $C_V = \frac{f}{2}R$.
Substituting $f = 3$,we get $C_V = \frac{3}{2}R$.

(D) For a diatomic gas,the degrees of freedom $(f)$ is $5$ at moderate temperatures.
Using the formula for molar specific heat at constant pressure,$C_p = (\frac{f}{2} + 1)R$.
Substituting $f = 5$,we get $C_p = (\frac{5}{2} + 1)R = \frac{7}{2}R$.
Thus,option $D$ is correct.

(D) For a diatomic gas,the molar specific heat at constant volume is given by the formula $C_V = \frac{f}{2}R$,where $f$ is the degrees of freedom.
For a diatomic gas at moderate temperatures,the degrees of freedom $f = 5$ ($3$ translational + $2$ rotational).
Thus,$C_V = \frac{5}{2}R$.
Using the value of the universal gas constant $R \approx 2 \, cal/mol \cdot K$,we get $C_V \approx \frac{5}{2} \times 2 = 5 \, cal/mol \cdot ^{\circ}C$.
Since all diatomic gases have the same degrees of freedom at these temperatures,their molar specific heat at constant volume is approximately equal to $5 \, cal/mol \cdot ^{\circ}C$.

(C) The molar specific heat at constant volume is given by the formula $C_V = \frac{R}{\gamma - 1}$.
Given that $C_V = \frac{3R}{2}$.
Equating the two,we get $\frac{R}{\gamma - 1} = \frac{3R}{2}$.
Dividing both sides by $R$,we have $\frac{1}{\gamma - 1} = \frac{3}{2}$.
Taking the reciprocal,$\gamma - 1 = \frac{2}{3}$.
Therefore,$\gamma = 1 + \frac{2}{3} = \frac{5}{3}$.
Calculating the value,$\gamma \approx 1.67$.

(A) According to Mayer's relation,for an ideal gas,the difference between the molar heat capacity at constant pressure $(C_P)$ and the molar heat capacity at constant volume $(C_V)$ is equal to the universal gas constant $(R)$.
When $C_P$ and $C_V$ are expressed in calories per mole per Kelvin $(cal \cdot mol^{-1} \cdot K^{-1})$ and $R$ is in Joules per mole per Kelvin $(J \cdot mol^{-1} \cdot K^{-1})$,we must divide $R$ by the mechanical equivalent of heat $(J)$ to maintain unit consistency.
Therefore,the relation is $C_P - C_V = \frac{R}{J}$.

(B) The change in internal energy $\Delta U$ for an ideal gas is given by the formula: $\Delta U = n C_V \Delta T$.
Given:
Number of moles $n = 1 \, mol$.
Molar specific heat at constant volume $C_V = 21.2 \, J/mol/^{\circ}C$.
Change in temperature $\Delta T = 1^{\circ}C$.
Substituting these values into the formula:
$\Delta U = 1 \times 21.2 \times 1 = 21.2 \, J$.
Therefore,the change in internal energy is $21.2 \, J$.

(D) For an ideal gas,the internal energy $U$ is given by:
$U = \frac{f}{2} RT$
where $f$ is the degree of freedom and $R$ is the universal gas constant.
Specific heat at constant volume is defined as $C_V = \frac{dU}{dT}$.
$C_V = \frac{d}{dT} (\frac{f}{2} RT) = \frac{f}{2} R$.
Since $f$ and $R$ are constants for an ideal gas,$C_V$ is independent of temperature $T$.
Similarly,$C_P = C_V + R$ is also independent of temperature.
Therefore,the specific heat of an ideal gas is independent of $T$.

(A) The formula for molar specific heat at constant volume is given by ${C_V} = \frac{f}{2}R$,where $f$ is the number of degrees of freedom.
For a monoatomic gas,the number of degrees of freedom is $f = 3$.
Substituting this value into the formula,we get ${C_V} = \frac{3}{2}R$.

(A) For an ideal gas,the relationship between molar specific heats is given by Mayer's relation: ${C_P} - {C_V} = R$.
Given the units are in $cal/gm-mole-K$,the universal gas constant $R \approx 2 \; cal/mol-K$.
For a monoatomic gas,${C_V} = \frac{3}{2}R = 3 \; cal/mol-K$ and ${C_P} = \frac{5}{2}R = 5 \; cal/mol-K$.
Checking the options:
Option $A$: ${C_P} - {C_V} = 5 - 3 = 2 = R$. This satisfies Mayer's relation.
Option $B$: ${C_P} - {C_V} = 6 - 4 = 2 = R$. While this satisfies Mayer's relation,${C_V} = 4$ is not a standard value for simple ideal gases.
Option $C$: ${C_P} < {C_V}$,which is physically impossible as ${C_P}$ must be greater than ${C_V}$.
Option $D$: ${C_P} - {C_V} = 4.2 - 3 = 1.2 \neq R$.
Thus,the set in option $A$ is the most reliable as it corresponds to the theoretical values for a monoatomic gas.

(A) The relation between molar specific heat $C_V$ and specific heat $c_V$ is given by $C_V = M \cdot c_V$,where $M$ is the molar mass.
Given $c_V = 0.075 \, kcal/kg-K = 0.075 \, cal/g-K$ and $C_V = 2.98 \, cal/mole-K$.
Calculating the molar mass $M$:
$M = \frac{C_V}{c_V} = \frac{2.98}{0.075} \approx 39.73 \, g/mole$.
The mass of a single atom is the molar mass divided by Avogadro's number $N_A$:
$m = \frac{M}{N_A} = \frac{39.73}{6.02 \times 10^{23}} \approx 6.60 \times 10^{-23} \, g$.

(B) We know that the molar specific heat at constant volume is given by ${C_V} = \frac{R}{J(\gamma - 1)}$.
Substituting $\gamma = 1.5$,we get ${C_V} = \frac{R}{J(1.5 - 1)} = \frac{R}{J(0.5)} = \frac{2R}{J}$.
Using Mayer's relation,${C_P} - {C_V} = \frac{R}{J}$,we have ${C_P} = {C_V} + \frac{R}{J}$.
Substituting the value of ${C_V}$,we get ${C_P} = \frac{2R}{J} + \frac{R}{J} = \frac{3R}{J}$.
Thus,the correct option is ${C_P} = \frac{3R}{J}$.

(A) By definition,specific heat at constant volume $(C_V)$ is the heat required to raise the temperature of $1 \text{ mole}$ of a gas by $1 \text{ K}$ while keeping the volume constant. In this case,no external work is done $(W = 0)$,so all the heat supplied is used to increase the internal energy of the gas.
However,at constant pressure $(C_P)$,when heat is supplied to the gas,it not only increases the internal energy but also performs external work $(W = P \Delta V)$ to expand the gas against the constant external pressure.
According to the first law of thermodynamics,$\Delta Q = \Delta U + W$. Since $C_P$ requires additional energy to perform this expansion work,more heat must be supplied to achieve the same temperature rise compared to the constant volume process. Therefore,$C_P > C_V$.

(C) The specific heat capacity of a gas is defined as $C = \frac{dQ}{m \cdot dT}$.
Since the heat capacity depends on the process (the path taken),for a given temperature change $dT$,the heat supplied $dQ$ can vary depending on the process.
For an isothermal process,$dT = 0$,so $C = \frac{dQ}{0} = \infty$.
For an adiabatic process,$dQ = 0$,so $C = \frac{0}{m \cdot dT} = 0$.
Therefore,the specific heat of a gas can take any value between $0$ and $\infty$ depending on the thermodynamic process.

(B) For a monoatomic gas,the degrees of freedom $f = 3$.
The molar specific heat at constant volume is given by ${C_V} = \frac{f}{2}R = \frac{3}{2}R$.
Using Mayer's relation,${C_P} - {C_V} = R$,we can find the molar specific heat at constant pressure ${C_P}$.
${C_P} = R + {C_V} = R + \frac{3}{2}R = \frac{5}{2}R$.

(C) The adiabatic index is given by $\gamma = 1 + \frac{2}{f}$,where $f$ is the degree of freedom.
Given $\gamma = 1.4$,we have $1.4 = 1 + \frac{2}{f}$.
This implies $\frac{2}{f} = 0.4$,so $f = \frac{2}{0.4} = 5$.
$A$ gas with $f = 5$ degrees of freedom is diatomic.
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 ${C_p} = {C_v} + R = \frac{5}{2}R + R = \frac{7}{2}R$.
Therefore,the gas is diatomic,${C_p} = \frac{7}{2}R$,and ${C_v} = \frac{5}{2}R$.

(D) According to Mayer's relation for an ideal gas,the difference between the molar specific heat at constant pressure $(C_P)$ and the molar specific heat at constant volume $(C_V)$ is equal to the universal gas constant $(R)$.
Mathematically,this is expressed as $C_P - C_V = R$.
Therefore,the formula $C_P - C_V = 2R$ is incorrect.

(B) For a diatomic gas,the number of degrees of freedom $f = 5$ ($3$ translational and $2$ rotational at room temperature).
The molar specific heat at constant volume is given by $C_V = \frac{f}{2}R = \frac{5}{2}R$.
The molar specific heat at constant pressure is given by $C_P = C_V + R = \frac{5}{2}R + R = \frac{7}{2}R$.
The ratio of specific heats $\gamma$ is defined as $\gamma = \frac{C_P}{C_V}$.
Substituting the values,$\gamma = \frac{7/2 R}{5/2 R} = \frac{7}{5} = 1.4$.

(D) According to Mayer's relation,for any ideal gas,the difference between the molar heat capacity at constant pressure $(C_p)$ and the molar heat capacity at constant volume $(C_v)$ is equal to the universal gas constant $(R)$.
Mathematically,$C_p - C_v = R$.
Since $R$ is a universal constant,it is the same for all ideal gases regardless of their molecular weight or composition.
For hydrogen gas,$C_p - C_v = a = R$.
For oxygen gas,$C_p - C_v = b = R$.
Therefore,$a = b$.

(D) Given,the difference between the two specific heats is $C_P - C_V = R = 4150 \ J/kg \ K$.
The ratio of specific heats is given by $\gamma = \frac{C_P}{C_V} = 1.4$.
We know that $C_P = \gamma C_V$.
Substituting this into the first equation: $\gamma C_V - C_V = R$.
$C_V(\gamma - 1) = R$.
$C_V = \frac{R}{\gamma - 1}$.
Substituting the values: $C_V = \frac{4150}{1.4 - 1} = \frac{4150}{0.4} = 10375 \ J/kg \ K$.

(A) The heat required to raise the temperature of $n$ moles of a gas by $\Delta T$ at constant volume is given by $(\Delta Q)_V = n C_V \Delta T$.
Given $n = 1 \text{ mole}$,$\Delta T = 1 \text{ K}$,the heat required is $(\Delta Q)_V = 1 \times C_V \times 1 = C_V$.
For a monoatomic gas,the molar specific heat at constant volume is $C_V = \frac{3}{2}R$.
Therefore,the required heat is $(\Delta Q)_V = \frac{3}{2}R$.

(B) For an ideal gas,the relationship between the molar specific heat at constant pressure $(C_P)$ and the molar specific heat at constant volume $(C_V)$ is given by Mayer's relation.
This relation is expressed as $C_P - C_V = R$,where $R$ is the universal gas constant.

(D) According to Mayer's relation,for any ideal gas,the difference between the molar specific heat at constant pressure $(C_P)$ and the molar specific heat at constant volume $(C_V)$ is equal to the universal gas constant $(R)$.
$C_P - C_V = R$.
The value of the universal gas constant $R$ is approximately $1.987 \, \text{cal/mol-K}$,which is often rounded to $1.99 \, \text{cal/mol-K}$.
Therefore,for hydrogen gas $(H_2)$,which behaves as an ideal gas under standard conditions,$C_P - C_V = 1.99 \, \text{cal/mol-K}$ is the correct statement.

(A) $NH_3$ is a non-linear polyatomic gas.
For a non-linear polyatomic molecule,the number of degrees of freedom is $f = 6$.
The ratio of specific heats $\gamma$ is given by the formula:
$\gamma = \frac{C_P}{C_V} = 1 + \frac{2}{f}$
Substituting $f = 6$ into the equation:
$\gamma = 1 + \frac{2}{6} = 1 + \frac{1}{3} = \frac{4}{3} \approx 1.33$.
Therefore,the ratio of specific heats for $NH_3$ is $1.33$.

(A) The heat supplied at constant volume is given by the formula: $(\Delta Q)_V = n C_V \Delta T$.
Since $n = 1 \, mol$,we have $(\Delta Q)_V = C_V \Delta T$.
We know that the molar heat capacity at constant volume is $C_V = \frac{f}{2} R$,where $f$ is the degree of freedom.
Thus,$\Delta T = \frac{(\Delta Q)_V}{C_V} = \frac{(\Delta Q)_V}{\frac{f}{2} R} = \frac{2(\Delta Q)_V}{f R}$.
Since $(\Delta Q)_V$ and $R$ are constant,$\Delta T \propto \frac{1}{f}$.
For a monoatomic gas,$f = 3$,and for a diatomic gas,$f = 5$.
Since $f_{\text{monoatomic}} < f_{\text{diatomic}}$,it follows that $\Delta T_{\text{monoatomic}} > \Delta T_{\text{diatomic}}$.
Therefore,the rise in temperature is more for the monoatomic gas.

(C) According to Mayer's relation,the relationship between the gas constant $R$ and the molar specific heats $C_P$ and $C_V$ is given by $C_P - C_V = \frac{R}{J}$.
Rearranging the formula to solve for $J$,we get $J = \frac{R}{C_P - C_V}$.
Given values are $R = 8.32 \ J/mol \cdot K$ and $C_P - C_V = 1.98 \ cal/mol \cdot K$.
Substituting these values into the equation:
$J = \frac{8.32}{1.98} \approx 4.20 \ J/cal$.

(B) The ratio of specific heats $\gamma$ is defined as the ratio of molar specific heat at constant pressure $C_p$ to molar specific heat at constant volume $C_v$,i.e.,$\gamma = \frac{C_p}{C_v}$.
According to the equipartition theorem,$C_v = \frac{f}{2}R$ and $C_p = C_v + R = (\frac{f}{2} + 1)R$.
Therefore,$\gamma = \frac{(\frac{f}{2} + 1)R}{\frac{f}{2}R} = \frac{\frac{f+2}{2}}{\frac{f}{2}} = 1 + \frac{2}{f}$.
Rearranging the equation: $\gamma - 1 = \frac{2}{f}$.
Solving for $f$,we get $f = \frac{2}{\gamma - 1}$.

(C) The ratio of specific heats $\gamma$ is defined as the ratio of molar specific heat at constant pressure $(C_P)$ to molar specific heat at constant volume $(C_V)$.
According to the law of equipartition of energy,the internal energy per mole is $U = \frac{n}{2}RT$.
Thus,$C_V = \frac{dU}{dT} = \frac{n}{2}R$.
Using Mayer's relation,$C_P = C_V + R = \frac{n}{2}R + R = R(1 + \frac{n}{2})$.
Therefore,$\gamma = \frac{C_P}{C_V} = \frac{R(1 + \frac{n}{2})}{\frac{n}{2}R} = \frac{1 + \frac{n}{2}}{\frac{n}{2}} = \frac{2}{n} + 1 = 1 + \frac{2}{n}$.

(B) The change in internal energy $\Delta U$ for an ideal gas at constant volume is given by $\Delta U = n C_V \Delta T$.
Using Mayer's relation,$C_V = C_P - R$.
Given $n = 5 \text{ moles}$,$C_P = 8 \text{ cal/mole } ^{\circ}C$,$R = 2 \text{ cal/mole } ^{\circ}C$,and $\Delta T = 20^{\circ}C - 10^{\circ}C = 10^{\circ}C$.
Calculating $C_V = 8 - 2 = 6 \text{ cal/mole } ^{\circ}C$.
Substituting the values: $\Delta U = 5 \times 6 \times 10 = 300 \text{ cal}$.

(C) For a diatomic gas,the molar heat capacity at constant pressure is given by $C_P = \frac{7}{2}R$.
Substituting this value into the expression $\frac{R}{C_P}$:
$\frac{R}{C_P} = \frac{R}{\frac{7}{2}R} = \frac{1}{\frac{7}{2}} = \frac{2}{7}$.
Thus,the correct value is $\frac{2}{7}$.

(B) Given: $\mu = 2 \, moles$,$(\Delta Q)_P = 70 \, cal$,$\Delta T = 35^{\circ}C - 30^{\circ}C = 5 \, K$,$R = 2 \, cal/mol \cdot K$.
At constant pressure,the heat supplied is given by $(\Delta Q)_P = \mu C_P \Delta T$.
Substituting the values: $70 = 2 \times C_P \times 5$.
$70 = 10 \times C_P$,which gives $C_P = 7 \, cal/mol \cdot K$.
Using the Mayer's relation: $C_P - C_V = R$.
$C_V = C_P - R = 7 - 2 = 5 \, cal/mol \cdot K$.
Now,at constant volume,the heat required is $(\Delta Q)_V = \mu C_V \Delta T$.
$(\Delta Q)_V = 2 \times 5 \times 5 = 50 \, cal$.

(B) For an ideal monoatomic gas,the molar heat capacity at constant volume is $C_v = \frac{3}{2}R$ and at constant pressure is $C_p = \frac{5}{2}R$.
Given $R = 2 \, \text{cal} \, \text{mol}^{-1} \text{K}^{-1}$,we have $C_v = \frac{3}{2} \times 2 = 3 \, \text{cal} \, \text{mol}^{-1} \text{K}^{-1}$ and $C_p = \frac{5}{2} \times 2 = 5 \, \text{cal} \, \text{mol}^{-1} \text{K}^{-1}$.
For a constant pressure process,the heat supplied is $Q_p = n C_p \Delta T$.
Given $Q_p = 40 \, \text{cal}$,$n = 1 \, \text{mole}$,and $\Delta T = 30^{\circ}C - 20^{\circ}C = 10 \, \text{K}$.
Checking the consistency: $Q_p = 1 \times 5 \times 10 = 50 \, \text{cal}$.
However,the problem states $40 \, \text{cal}$ is required. This implies the gas might not be perfectly monoatomic or the provided $R$ is for calculation purposes. Using the ratio $C_v/C_p = 1/\gamma$ where $\gamma = 5/3$ for monoatomic gas:
$C_v = C_p / \gamma = (C_p \times 3) / 5$.
From $Q_p = n C_p \Delta T$,we have $40 = 1 \times C_p \times 10$,so $C_p = 4 \, \text{cal} \, \text{mol}^{-1} \text{K}^{-1}$.
Then $C_v = 4 \times (3/5) = 2.4 \, \text{cal} \, \text{mol}^{-1} \text{K}^{-1}$.
For constant volume,$Q_v = n C_v \Delta T = 1 \times 2.4 \times 10 = 24 \, \text{cal}$.

(C) The ratio of specific heat at constant pressure to specific heat at constant volume is given by the adiabatic index $\gamma = C_p / C_v$.
For a polyatomic gas like $NH_3$ (ammonia),which has non-linear molecules,the degrees of freedom $f = 6$.
The adiabatic index is calculated as $\gamma = 1 + (2/f)$.
Substituting $f = 6$,we get $\gamma = 1 + (2/6) = 1 + 0.33 = 1.33$.
However,for real polyatomic gases,the vibrational degrees of freedom contribute,making $\gamma$ typically lower than the ideal value of $1.33$.
Among the given options,$1.28$ is the most accurate value for $NH_3$ at room temperature.

(C) Given: $C_p = 7.2 \, cal/mol \cdot K$,$R = 8.3 \, J/mol \cdot K$. Since $1 \, cal \approx 4.18 \, J$,we convert $R$ to $cal$: $R = 8.3 / 4.18 \approx 2 \, cal/mol \cdot K$.
Molar heat capacity at constant volume is given by $C_v = C_p - R$.
$C_v = 7.2 - 2 = 5.2 \, cal/mol \cdot K \approx 5 \, cal/mol \cdot K$.
Number of moles $n = 5 \, mol$.
Change in temperature $\Delta T = 20^\circ C - 10^\circ C = 10 \, K$.
The heat required at constant volume is $\Delta Q = n C_v \Delta T$.
$\Delta Q = 5 \times 5 \times 10 = 250 \, cal$.

(D) The heat required at constant pressure is given by $(\Delta Q)_P = n C_P \Delta T$.
Given $n = 1 \, mol$,$(\Delta Q)_P = 207 \, J$,and $\Delta T = 10 \, K$.
$207 = 1 \times C_P \times 10 \implies C_P = 20.7 \, J/mol \cdot K$.
Using Mayer's relation,$C_P - C_V = R$,we find $C_V = C_P - R$.
$C_V = 20.7 - 8.3 = 12.4 \, J/mol \cdot K$.
The heat required at constant volume is $(\Delta Q)_V = n C_V \Delta T$.
$(\Delta Q)_V = 1 \times 12.4 \times 10 = 124 \, J$.

(C) For a monoatomic gas,the molar specific heat at constant volume $(C_v)$ is given by the formula $C_v = \frac{f}{2}R$,where $f$ is the degrees of freedom. For a monoatomic gas,$f = 3$. Therefore,$C_v = \frac{3}{2}R$. This value is a constant and does not depend on the temperature $(T)$. Thus,the graph of $C_v$ versus $T$ is a horizontal straight line at the value $\frac{3}{2}R$.

(C) The change in internal energy $\Delta U$ for an ideal gas is given by the formula $\Delta U = n C_V \Delta T$.
Since $C_V = R/(\gamma - 1)$,we have $\Delta U = n (R/(\gamma - 1)) \Delta T$.
Using the ideal gas law $pV = nRT$,at constant pressure $p$,we have $p \Delta V = nR \Delta T$.
Substituting $n R \Delta T = p \Delta V$ into the equation,we get $\Delta U = p \Delta V / (\gamma - 1)$.
Given the volume changes from $V$ to $2V$,the change in volume is $\Delta V = 2V - V = V$.
Therefore,$\Delta U = pV / (\gamma - 1)$.

(D) The fraction of supplied heat energy that increases the internal energy is given by the ratio of the change in internal energy to the heat supplied at constant pressure.
$f = \frac{\Delta U}{(\Delta Q)_P} = \frac{(\Delta Q)_V}{(\Delta Q)_P} = \frac{\mu C_V \Delta T}{\mu C_P \Delta T} = \frac{C_V}{C_P} = \frac{1}{\gamma}$.
For an ideal diatomic gas,the adiabatic index $\gamma = \frac{C_P}{C_V} = \frac{7}{5}$.
Therefore,the fraction $f = \frac{1}{7/5} = \frac{5}{7}$.

(B) For an ideal monoatomic gas,the molar heat capacity at constant pressure is $C_P = \frac{5}{2}R$ and at constant volume is $C_V = \frac{3}{2}R$.
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 fraction of heat energy that increases the internal energy is $\frac{\Delta U}{\Delta Q} = \frac{n C_V \Delta T}{n C_P \Delta T} = \frac{C_V}{C_P}$.
Substituting the values,we get $\frac{\Delta U}{\Delta Q} = \frac{3/2 R}{5/2 R} = \frac{3}{5} = 0.6$.

(C) Given,$\frac{R}{C_V} = 0.67$.
We know that $R = C_P - C_V$.
Substituting this in the equation: $\frac{C_P - C_V}{C_V} = 0.67$.
This simplifies to $\frac{C_P}{C_V} - 1 = 0.67$.
Since $\gamma = \frac{C_P}{C_V}$,we have $\gamma - 1 = 0.67$,which gives $\gamma = 1.67$.
For a monoatomic gas,$\gamma = \frac{5}{3} \approx 1.67$.
Therefore,the gas is monoatomic.

(A) We know that for an ideal gas,the relation between molar specific heats is given by Mayer's relation: $C_P - C_V = R$.
Given that $\gamma = \frac{C_P}{C_V}$,we can write $C_P = \gamma C_V$.
Substituting this into the 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: $n = 1$ $mol$,$\Delta T = 10$ $K$,$Q_P = 207$ $J$.
At constant pressure,$Q_P = n C_P \Delta T$.
$207 = 1 \times C_P \times 10 \implies C_P = 20.7$ $J/mol$ $K$.
We know that $C_P - C_V = R$.
$C_V = C_P - R = 20.7 - 8.3 = 12.4$ $J/mol$ $K$.
At constant volume,the heat required is $Q_V = n C_V \Delta T$.
$Q_V = 1 \times 12.4 \times 10 = 124$ $J$.

(B) Given,the molar specific heat at constant pressure is $C_P = \frac{7}{2}R$.
We know the relation between molar specific heats for an ideal gas is $C_P - C_V = R$.
Substituting the value of $C_P$,we get $\frac{7}{2}R - C_V = R$.
Rearranging for $C_V$,we get $C_V = \frac{7}{2}R - R = \frac{5}{2}R$.
The ratio of molar specific heat at constant pressure to constant volume is $\gamma = \frac{C_P}{C_V}$.
Substituting the values,$\gamma = \frac{(7/2)R}{(5/2)R} = \frac{7}{5}$.

(B) For a monoatomic gas,the molar heat capacity at constant volume is $C_V = \frac{3}{2}R$ and at constant pressure is $C_P = \frac{5}{2}R$.
When heat $\Delta Q$ is supplied at constant pressure,the total heat energy is given by $\Delta Q = \mu C_P \Delta T$.
The change in internal energy is given by $\Delta U = \mu C_V \Delta T$.
The fraction of heat energy used to increase internal energy is $\frac{\Delta U}{\Delta Q} = \frac{\mu C_V \Delta T}{\mu C_P \Delta T} = \frac{C_V}{C_P}$.
Substituting the values,we get $\frac{\Delta U}{\Delta Q} = \frac{3/2 R}{5/2 R} = \frac{3}{5}$.

(D) For an ideal gas,the heat required to change the temperature at constant volume is given by $\Delta Q = \mu C_V \Delta T$.
Helium is a monoatomic gas,so its degrees of freedom $f = 3$.
The molar specific heat at constant volume is $C_V = \frac{fR}{2} = \frac{3R}{2}$.
The number of moles $\mu = \frac{\text{mass}}{\text{molar mass}} = \frac{1 \ g}{4 \ g/mol} = \frac{1}{4} \ mol$.
Substituting these values into the formula:
$\Delta Q = \left( \frac{1}{4} \right) \left( \frac{3R}{2} \right) (T_2 - T_1) = \frac{3R}{8} (T_2 - T_1)$.
Since the universal gas constant $R = N_A k_B$,we substitute $R$ to get:
$\Delta Q = \frac{3}{8} N_A k_B (T_2 - T_1)$.