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

(B) The heat supplied at constant pressure is given by $Q_p = n C_p \Delta T$.
The change in internal energy is given by $\Delta U = n C_v \Delta T$.
The fraction of heat energy used to increase the internal energy is $f = \frac{\Delta U}{Q_p} = \frac{n C_v \Delta T}{n C_p \Delta T} = \frac{C_v}{C_p} = \frac{1}{\gamma}$.
For an ideal non-linear triatomic gas,the degrees of freedom $f_{deg} = 6$.
The adiabatic index $\gamma = 1 + \frac{2}{f_{deg}} = 1 + \frac{2}{6} = 1 + \frac{1}{3} = \frac{4}{3}$.
Therefore,the fraction $f = \frac{1}{4/3} = \frac{3}{4} = 0.75$.

(D) According to the law of equipartition of energy, each degree of freedom contributes $\frac{1}{2} k_B T$ to the average energy of a molecule.
For a solid, each atom acts as a 3D harmonic oscillator, having $3$ degrees of freedom for kinetic energy and $3$ degrees of freedom for potential energy, totaling $6$ degrees of freedom.
The average energy per atom is $U = 6 \times \frac{1}{2} k_B T = 3 k_B T$.
For $1 \, \text{mole}$ of substance, the internal energy is $U_m = 3 R T$.
The molar specific heat capacity is $C_v = \frac{dU_m}{dT} = 3 R$.
Given $R = 8.314 \, J \, mol^{-1} \, K^{-1}$ and atomic weight $M = 27 \times 10^{-3} \, kg/mol$, the specific heat capacity $c$ is given by $c = \frac{C_v}{M} = \frac{3 R}{M}$.
Substituting the values: $c = \frac{3 \times 8.314}{27 \times 10^{-3}} \approx \frac{24.942}{0.027} \approx 923.77 \, J \, kg^{-1} \, K^{-1}$.
Rounding to the nearest provided option, we get $925 \, J \, kg^{-1} \, K^{-1}$.

(A) For an isobaric process,the heat supplied is given by $Q = n C_p \Delta T$,where $n$ is the number of moles and $C_p$ is the molar heat capacity at constant pressure.
From the graph,the slope of the line is $\frac{Q}{\Delta T} = n C_p$.
Since the number of moles $n$ is the same for all gases,the slope is directly proportional to $C_p$.
The molar heat capacity at constant pressure is $C_p = C_v + R = \left( \frac{f}{2} + 1 \right) R$,where $f$ is the degree of freedom.
For monoatomic gas $(M)$,$f = 3$,so $C_p = (1.5 + 1) R = 2.5 R$.
For diatomic gas $(D)$,$f = 5$,so $C_p = (2.5 + 1) R = 3.5 R$.
For polyatomic gas $(P)$,$f = 6$,so $C_p = (3 + 1) R = 4 R$.
Thus,$C_p(P) > C_p(D) > C_p(M)$.
Since the slope is proportional to $C_p$,the slopes follow the order: $\text{slope}(a) > \text{slope}(b) > \text{slope}(c)$.
Therefore,line $a$ corresponds to $P$,line $b$ corresponds to $D$,and line $c$ corresponds to $M$.

(C) At $STP$,the molar volume of an ideal gas is $22.4 \, L/mol$.
The number of moles $n$ of helium gas is $n = \frac{67.2 \, L}{22.4 \, L/mol} = 3 \, mol$.
Helium is a monoatomic gas,so its molar heat capacity at constant volume is $C_v = \frac{3}{2}R$.
The heat required is given by $\Delta Q = n C_v \Delta T$.
Substituting the values: $\Delta Q = 3 \times \left(\frac{3}{2} \times 8.31 \right) \times 20$.
$\Delta Q = 3 \times 1.5 \times 8.31 \times 20 = 4.5 \times 166.2 = 747.9 \, J$.
Rounding to the nearest integer,we get $\Delta Q \approx 748 \, J$.

(C) For a diatomic gas of rigid molecules,the molar heat capacity at constant volume is $C_V = \frac{5}{2} R$.
Given that heat $Q$ is supplied at constant volume,we have $Q = n C_V \Delta T = n (\frac{5}{2} R) \Delta T$.
For the same change in temperature $\Delta T$ at constant pressure,the heat required is $Q' = n C_P \Delta T$,where $C_P = \frac{7}{2} R$.
Taking the ratio,we get $\frac{Q'}{Q} = \frac{n C_P \Delta T}{n C_V \Delta T} = \frac{C_P}{C_V} = \frac{7/2 R}{5/2 R} = \frac{7}{5}$.
Therefore,$Q' = \frac{7}{5} Q$.

(C) For an ideal gas,the change in internal energy is given by $\Delta U = n C_V \Delta T$.
Using $C_V = \frac{R}{\gamma - 1}$,we get $\Delta U = n \left( \frac{R}{\gamma - 1} \right) (T_f - T_i)$.
Since $PV = nRT$,we have $nRT_f = P_f V_f$ and $nRT_i = P_i V_i$.
Substituting these,$\Delta U = \frac{P_f V_f - P_i V_i}{\gamma - 1}$.
Comparing this with option $C$,we see that $\frac{P_f V_f - P_i V_i}{1 - \gamma} = - \left( \frac{P_f V_f - P_i V_i}{\gamma - 1} \right)$,which is incorrect. Thus,option $C$ is the incorrect relation.

(B) For an ideal mono-atomic gas,the change in internal energy is given by $dU = nC_v dT$,where $C_v = \frac{3}{2}R$.
Given the process condition $dQ = 3dU$.
We know that the molar heat capacity $C$ is defined as $C = \frac{dQ}{n dT}$.
Substituting the given condition: $C = \frac{3dU}{n dT} = \frac{3(nC_v dT)}{n dT} = 3C_v$.
Substituting the value of $C_v$ for a mono-atomic gas: $C = 3 \times \left(\frac{3}{2}R\right) = \frac{9}{2}R = 4.5R$.

(D) The given equation is $\left( \frac{T^7}{P^2} \right)^{1/5} = \text{constant}$.
This can be rewritten as $T^{7/5} P^{-2/5} = \text{constant}$.
For an adiabatic process,the equation relating temperature $T$ and pressure $P$ is $T^{\gamma} P^{1-\gamma} = \text{constant}$.
Comparing the exponents of $T$ and $P$,we have $\gamma = 7/5$ and $1 - \gamma = -2/5$.
Solving for $\gamma$,we get $\gamma = 1 + 2/5 = 7/5 = 1.4$.
The molar specific heat at constant volume $C_v$ is given by the formula $C_v = \frac{R}{\gamma - 1}$.
Substituting $\gamma = 7/5$,we get $C_v = \frac{R}{7/5 - 1} = \frac{R}{2/5} = 2.5 R$.

(B) According to Mayer's relation for specific heat capacities,the difference between the specific heat at constant pressure $(c_p)$ and the specific heat at constant volume $(c_v)$ is given by $c_p - c_v = R/M$,where $R$ is the universal gas constant and $M$ is the molar mass of the gas.
For nitrogen gas $(N_2)$,the molar mass $M = 28 \ g/mol$.
Substituting this value into the relation,we get $c_p - c_v = R/28$.

(D) The ratio of specific heats is defined as $\gamma = \frac{C_P}{C_V}$.
From Mayer's relation,we know that $C_P - C_V = R$,which implies $C_V = C_P - R$.
Substituting this into the ratio,we get $\gamma = \frac{C_P}{C_P - R}$.
Also,from $C_P - C_V = R$,we can write $C_P = C_V + R$,so $\gamma = \frac{C_V + R}{C_V} = 1 + \frac{R}{C_V}$.
Furthermore,$\gamma = \frac{C_P}{C_P - R} = \frac{1}{\frac{C_P - R}{C_P}} = \frac{1}{1 - \frac{R}{C_P}}$.
Since all three expressions are equivalent,the correct option is $D$.

(D) Since the volume of the gas remains constant,the heat energy required is given by $\Delta Q = n C_v \Delta T$.
Here,the number of moles $n$ for $1\,g$ of Helium (molar mass $M = 4\,g/mol$) is $n = \frac{1}{4}$.
For a monoatomic gas like Helium,the molar heat capacity at constant volume is $C_v = \frac{3}{2} R$.
The change in temperature is $\Delta T = T_2 - T_1$.
Substituting these values,we get $\Delta Q = \frac{1}{4} \times \left( \frac{3}{2} R \right) \times (T_2 - T_1) = \frac{3}{8} R (T_2 - T_1)$.
Using the relation $R = N_a k_B$,where $N_a$ is Avogadro's number and $k_B$ is the Boltzmann constant,we get $\Delta Q = \frac{3}{8} N_a k_B (T_2 - T_1)$.
Thus,the correct option is $D$.

(B) The specific heat of a gas is defined as the amount of heat required to increase the temperature of $1 \text{ mole}$ of the gas by $1 \text{ K}$ (or $1^{\circ}C$).
During different thermodynamic processes,the work done by the gas and the change in internal energy vary.
Since the specific heat $C$ is given by $C = \frac{dQ}{dT} = \frac{dU}{dT} + \frac{dW}{dT}$,and the work done $dW$ depends on the path (process) taken,the value of $C$ is not fixed.
Because there are an infinite number of possible thermodynamic processes,a gas can have an infinite number of specific heats.
Therefore,option $B$ is correct.

(A) If $f$ is the degree of freedom,then the ratio $\gamma = \frac{C_p}{C_v}$ is given by $\gamma = 1 + \frac{2}{f}$.
For a monoatomic gas,$f = 3$.
Therefore,$\gamma_{\text{mono}} = 1 + \frac{2}{3} = \frac{5}{3} \approx 1.67$.
For a diatomic gas,$f = 5$.
Therefore,$\gamma_{\text{dia}} = 1 + \frac{2}{5} = \frac{7}{5} = 1.4$.
Since $1.4 < 1.67$,the ratio for a diatomic gas is indeed less than that for a monoatomic gas.
The reason is correct because the degree of freedom $f$ is higher for diatomic molecules $(f=5)$ than for monoatomic atoms $(f=3)$,which leads to a smaller value of $\gamma$.

(C) According to Mayer's relation,$C_p - C_v = R$. Since $R > 0$,it follows that $C_p > C_v$. This is because when a gas is heated at constant pressure,it expands and does work against external pressure. Therefore,extra heat is required to perform this work in addition to the heat required to increase the internal energy. Conversely,at constant volume,no work is done by the gas,so all supplied heat is used only to increase the internal energy. Thus,the Assertion is correct,but the Reason states the opposite (that extra heat is needed at constant volume),making the Reason incorrect.

(N/A) The gases listed in the table are diatomic. Besides the translational degrees of freedom,they possess rotational degrees of freedom.
Heat supplied to the gas increases the average energy of all these modes of motion. Consequently,the molar specific heat of diatomic gases is higher than that of monatomic gases.
If only translational and rotational modes are considered,the theoretical molar specific heat of a diatomic gas is given by $C_v = \frac{5}{2} R$.
Using $R \approx 1.98 \; cal\, mol^{-1}\, K^{-1}$,we get $C_v = 2.5 \times 1.98 = 4.95 \; cal\, mol^{-1}\, K^{-1}$.
Most gases in the table show values close to $4.95 \; cal\, mol^{-1}\, K^{-1}$. However,chlorine has a significantly higher value $(6.17 \; cal\, mol^{-1}\, K^{-1})$. This indicates that at room temperature,chlorine molecules also possess vibrational degrees of freedom in addition to translational and rotational modes,which contribute to the total internal energy and thus increase the molar specific heat.

(B) Mass of nitrogen,$m = 2.0 \times 10^{-2} \; kg = 20 \; g$.
Rise in temperature,$\Delta T = 45 \; ^{\circ}C$.
Molecular mass of $N_{2}$,$M = 28 \; g/mol$.
Universal gas constant,$R = 8.3 \; J \; mol^{-1} K^{-1}$.
Number of moles,$n = \frac{m}{M} = \frac{20}{28} = \frac{5}{7} \; mol$.
For a diatomic gas like nitrogen,the molar specific heat at constant pressure is $C_{p} = \frac{7}{2} R$.
$C_{p} = \frac{7}{2} \times 8.3 = 29.05 \; J \; mol^{-1} K^{-1}$.
The total amount of heat supplied at constant pressure is given by $\Delta Q = n C_{p} \Delta T$.
$\Delta Q = \left( \frac{5}{7} \right) \times 29.05 \times 45$.
$\Delta Q = \frac{5}{7} \times \left( \frac{7}{2} \times 8.3 \right) \times 45 = \frac{5}{2} \times 8.3 \times 45$.
$\Delta Q = 2.5 \times 373.5 = 933.75 \; J$.
Rounding to the nearest integer,the heat supplied is $933 \; J$.

(N/A) If the amount of substance is specified in terms of moles $\mu$,instead of mass $m$ in $kg$,we can define heat capacity per mole of the substance by:
$C = \frac{1}{\mu} \frac{\Delta Q}{\Delta T}$
where $C$ is known as the molar specific heat capacity of the substance. $C$ depends on the nature of the substance and its temperature.
The $SI$ unit of molar specific heat capacity is $J \cdot mol^{-1} \cdot K^{-1}$.
We can define molar specific heat for gases in two ways:
$(i)$ Molar specific heat at constant pressure $(C_P)$: If the gas is held under constant pressure during the heat transfer,the molar specific heat capacity is called the molar specific heat capacity at constant pressure,denoted by $C_P$.
$(ii)$ Molar specific heat at constant volume $(C_V)$: If the volume of the gas is maintained constant during the heat transfer,the corresponding molar specific heat capacity is called the molar specific heat capacity at constant volume,denoted by $C_V$.

(N/A) $1$. Molar specific heat at constant volume $(C_V)$: It is defined as the amount of heat required to raise the temperature of $1 \text{ mole}$ of a gas by $1 \text{ K}$ (or $1^{\circ}\text{C}$) while keeping its volume constant.
$2$. Molar specific heat at constant pressure $(C_P)$: It is defined as the amount of heat required to raise the temperature of $1 \text{ mole}$ of a gas by $1 \text{ K}$ (or $1^{\circ}\text{C}$) while keeping its pressure constant.

(C) The law of equipartition of energy can be used to predict the molar specific heat capacities of solids. Consider a solid consisting of $N$ atoms,each vibrating about its mean position.
An oscillator in one dimension has an average energy $= 2 \times \frac{1}{2} k_{B} T = k_{B} T$,where $k_{B}$ is the Boltzmann constant.
In three dimensions,the average energy $= 3 k_{B} T$.
Therefore,for one mole of a solid,the total internal energy $U$ is the average energy multiplied by the number of atoms in one mole $(N_{A})$:
$U = 3 k_{B} T \times N_{A}$
Since $k_{B} N_{A} = R$ (the universal gas constant),we have:
$U = 3 RT$
To find the molar specific heat capacity $C$,we differentiate $U$ with respect to temperature $T$:
$C = \frac{dU}{dT} = \frac{d}{dT}(3 RT) = 3R$
According to the first law of thermodynamics,$\Delta Q = \Delta U + \Delta W$. For a solid,the change in volume $\Delta V$ is negligible,so the work done $\Delta W = P \Delta V \approx 0$. Thus,$\Delta Q = \Delta U$. Therefore,the molar specific heat capacity of a solid is $C = 3R$.

(N/A) For gases,the specific heat capacity is defined as the molar specific heat capacity.
$1$. Molar specific heat capacity at constant volume $(C_{V})$: It is the amount of heat required to raise the temperature of $1$ mole of a gas by $1 \ K$ (or $1^{\circ}C$) while keeping the volume constant.
$2$. Molar specific heat capacity at constant pressure $(C_{P})$: It is the amount of heat required to raise the temperature of $1$ mole of a gas by $1 \ K$ (or $1^{\circ}C$) while keeping the pressure constant.
The relation between these two molar specific heat capacities is given by Mayer's relation: $C_{P} - C_{V} = R$,where $R$ is the universal gas constant.

(N/A) For $1$ mole of an ideal gas,the first law of thermodynamics is $\Delta Q = \Delta U + \Delta W$.
At constant volume,$\Delta W = 0$,so $\Delta Q = \Delta U$. Since $\Delta U = C_V \Delta T$,we have $C_V = \frac{\Delta U}{\Delta T}$.
For an ideal gas,the internal energy $U$ depends only on temperature,so $\Delta U = C_V \Delta T$ holds for any process.
At constant pressure,$\Delta Q = \Delta U + P \Delta V$. Dividing by $\Delta T$,we get $\frac{\Delta Q}{\Delta T} = \frac{\Delta U}{\Delta T} + P \frac{\Delta V}{\Delta T}$.
This gives $C_P = C_V + P \frac{\Delta V}{\Delta T}$.
From the ideal gas equation for $1$ mole,$PV = RT$. Differentiating at constant pressure,$P \Delta V = R \Delta T$,which implies $P \frac{\Delta V}{\Delta T} = R$.
Substituting this into the expression for $C_P$,we get $C_P = C_V + R$.
Therefore,the relation is $C_P - C_V = R$.

(B) The relation $C_P - C_V = R$ holds true for all ideal gases,where $R$ is the universal gas constant.
However,the ratio $\gamma = \frac{C_P}{C_V}$ depends on the atomicity of the gas.
For monoatomic gases,$\gamma = 1.67$.
For diatomic gases,$\gamma = 1.40$.
For polyatomic gases,$\gamma$ takes different values.
Since $\gamma$ varies based on the degrees of freedom of the gas molecules,it is not a constant for all gases.

(A) Yes,we can use the degree of freedom $(f)$ to determine the molar specific heat capacity of a gas.
According to the law of equipartition of energy,the energy associated with each degree of freedom per molecule is $\frac{1}{2} k_{B} T$.
For a gas with $f$ degrees of freedom,the total internal energy $(U)$ for $1$ mole of gas is given by:
$U = f \times \left( \frac{1}{2} k_{B} T \right) \times N_{A} = \frac{f}{2} RT$ (since $k_{B} N_{A} = R$)
The molar specific heat at constant volume $(C_{V})$ is defined as the rate of change of internal energy with respect to temperature:
$C_{V} = \frac{dU}{dT} = \frac{d}{dT} \left( \frac{f}{2} RT \right) = \frac{f}{2} R$
Using Mayer's relation,the molar specific heat at constant pressure $(C_{P})$ is:
$C_{P} = C_{V} + R = \frac{f}{2} R + R = \left( \frac{f}{2} + 1 \right) R$
The ratio of specific heats $(\gamma)$ is:
$\gamma = \frac{C_{P}}{C_{V}} = \frac{(\frac{f}{2} + 1)R}{\frac{f}{2}R} = 1 + \frac{2}{f}$

(D) For a diatomic rigid rotator gas,the degree of freedom $(f)$ is $5$,and the energy associated with each degree of freedom is $\frac{1}{2} k_{B} T$.
From the law of equipartition of energy,the total internal energy $(U)$ of $1 \text{ mole}$ of gas is:
$U = 5 \times \frac{1}{2} k_{B} T \times N_{A}$
$U = \frac{5}{2} (k_{B} N_{A}) T$
Since $k_{B} N_{A} = R$,we have:
$U = \frac{5}{2} RT \quad \dots(1)$
The molar specific heat at constant volume $(C_{V})$ is given by:
$C_{V} = \frac{dU}{dT} \quad \dots(2)$
Substituting equation $(1)$ into $(2)$:
$C_{V} = \frac{d}{dT} \left[ \frac{5}{2} RT \right] = \frac{5}{2} R$
Using Mayer's relation for molar specific heat at constant pressure $(C_{P})$:
$C_{P} - C_{V} = R$
$C_{P} = C_{V} + R = \frac{5}{2} R + R = \frac{7}{2} R$
Therefore,the ratio $\gamma = \frac{C_{P}}{C_{V}}$ is:
$\frac{C_{P}}{C_{V}} = \frac{\frac{7}{2} R}{\frac{5}{2} R} = \frac{7}{5} = 1.4$

(D) For a non-linear triatomic gas,the degrees of freedom $(f)$ is $6$ ($3$ translational + $3$ rotational). However,if the question implies a general triatomic gas at higher temperatures where vibrational modes are active,$f$ can be $7$. Assuming the standard case for a non-linear triatomic gas at room temperature,$f = 6$.
$C_V = \frac{f}{2} R = \frac{6}{2} R = 3R$
Using Mayer's relation,$C_P = C_V + R = 3R + R = 4R$
Therefore,$\gamma = \frac{C_P}{C_V} = \frac{4R}{3R} = 1.33$.
If the gas is treated as having $f=7$ (including vibrational modes),then $C_V = \frac{7}{2}R$ and $C_P = \frac{9}{2}R$,giving $\gamma = \frac{9}{7} \approx 1.28$.

(D) polyatomic gas has $3$ translational,$3$ rotational,and $f$ vibrational degrees of freedom. Thus,the total degree of freedom is $f_{total} = (3 + 3 + f) = (6 + f)$.
The internal energy $U$ is given by the law of equipartition of energy:
$U = \frac{f_{total}}{2} RT = \frac{6+f}{2} RT$.
The molar specific heat at constant volume is $C_{V} = \frac{dU}{dT} = \frac{6+f}{2} R$.
Using Mayer's relation,$C_{P} = C_{V} + R = \left(\frac{6+f}{2} + 1\right) R = \left(\frac{8+f}{2}\right) R$.
Therefore,the ratio $\gamma = \frac{C_{P}}{C_{V}} = \frac{\frac{8+f}{2} R}{\frac{6+f}{2} R} = \frac{8+f}{6+f}$.

(D) For a gas with $f$ degrees of freedom,the internal energy $U$ of $1 \text{ mole}$ of gas is given by:
$U = f \times \frac{1}{2} k_B T \times N_A = \frac{1}{2} f RT$ (since $k_B N_A = R$).
From the definition of molar specific heat at constant volume:
$C_V = \frac{dU}{dT} = \frac{d}{dT} \left( \frac{1}{2} f RT \right) = \frac{1}{2} f R$.
Using Mayer's relation for molar specific heat at constant pressure:
$C_P = C_V + R = \frac{1}{2} f R + R = \left( \frac{f}{2} + 1 \right) R$.
The adiabatic index $\gamma$ is defined as the ratio of specific heats:
$\gamma = \frac{C_P}{C_V} = \frac{(\frac{f}{2} + 1) R}{\frac{1}{2} f R}$.
Simplifying the expression:
$\gamma = \frac{f + 2}{f} = 1 + \frac{2}{f}$.

(N/A) If we consider a water molecule as a solid,it consists of three atoms ($2$ hydrogen and $1$ oxygen) vibrating about their mean positions.
From the equipartition of energy,the energy associated with one molecule of water is:
$= 2 \times \frac{1}{2} k_{B} T = k_{B} T$ (for kinetic energy).
Considering the vibrational energy in three dimensions (each atom has $3$ degrees of freedom for vibration,contributing $k_{B} T$ per atom):
Total energy of one molecule of water $= 3 \times k_{B} T = 3 k_{B} T$ (for kinetic part) and $3 \times k_{B} T$ (for potential part).
Total energy of one molecule of water $= 3 \times (k_{B} T + k_{B} T) = 6 k_{B} T$ (simplified model for solid-like behavior).
Total energy of one mole of water:
$U = 6 k_{B} T \times N_{A} = 6 RT$.
Molar specific heat of water:
$C = \frac{dU}{dT} = \frac{d}{dT}(6 RT) = 6R$.
$C = 6 \times 8.31 = 49.86 \text{ J mol}^{-1} \text{ K}^{-1}$.
(Note: The actual specific heat of liquid water is much higher due to hydrogen bonding,which is not accounted for in simple kinetic theory models.)

(N/A) According to the law of equipartition of energy,each degree of freedom contributes $\frac{1}{2} k_B T$ to the internal energy of a system.
For a gas with $f$ degrees of freedom,the internal energy $U$ is given by $U = f \cdot \frac{1}{2} n R T$.
The molar specific heat at constant volume is $C_V = \frac{dU}{dT} = \frac{f}{2} R$.
The molar specific heat at constant pressure is $C_P = C_V + R = (\frac{f}{2} + 1) R$.
This classical prediction assumes that all degrees of freedom are active at all temperatures.
However,experimental observations show that specific heat varies with temperature,approaching zero as $T \to 0 \ K$,which indicates that degrees of freedom become 'frozen' or ineffective at low temperatures.
This limitation of classical mechanics is explained by quantum mechanics,where a minimum energy threshold is required to excite a degree of freedom.

(A) For a monoatomic gas,the degrees of freedom $(f)$ is $3$.
The molar specific heat at constant volume is given by $C_V = \frac{f}{2}R = \frac{3}{2}R$.
The molar specific heat at constant pressure is given by $C_P = C_V + R = \frac{3}{2}R + R = \frac{5}{2}R$.
The ratio of specific heats $\gamma = \frac{C_P}{C_V}$ is calculated as:
$\gamma = \frac{\frac{5}{2}R}{\frac{3}{2}R} = \frac{5}{3} \approx 1.67$.

(B) The ratio of molar specific heat at constant pressure $(C_P)$ to molar specific heat at constant volume $(C_V)$ is known as the adiabatic index or the ratio of specific heats,denoted by $\gamma$.
For an ideal gas,$\gamma = \frac{C_P}{C_V} = 1 + \frac{2}{f}$,where $f$ is the number of degrees of freedom.
For a diatomic gas at room temperature,the number of degrees of freedom $f = 5$ ($3$ translational + $2$ rotational).
Substituting the value of $f$ into the formula: $\gamma = 1 + \frac{2}{5} = 1 + 0.4 = 1.40$.
Therefore,the ratio for a diatomic gas is $1.40$.

(A) For a polyatomic gas,the degrees of freedom $(f)$ is typically $6$ (assuming non-linear molecules with rotational and vibrational modes ignored at moderate temperatures).
The adiabatic index $\gamma$ is given by the formula $\gamma = 1 + \frac{2}{f}$.
Substituting $f = 6$ into the formula:
$\gamma = 1 + \frac{2}{6} = 1 + \frac{1}{3} = 1.33$.
Therefore,the value of $\gamma$ for a polyatomic gas is approximately $1.33$.

(A) For a monoatomic gas, the degrees of freedom $f = 3$. The molar heat capacity at constant volume is ${C_V} = \frac{f}{2}R = \frac{3}{2}R$. Using Mayer's relation, ${C_P} = {C_V} + R = \frac{3}{2}R + R = \frac{5}{2}R$. Thus, $(a)$ matches with $(ii)$.
For a diatomic gas with vibration, the degrees of freedom $f = 7$ ($3$ translational + $2$ rotational + $2$ vibrational). The molar heat capacity at constant volume is ${C_V} = \frac{f}{2}R = \frac{7}{2}R$. Using Mayer's relation, ${C_P} = {C_V} + R = \frac{7}{2}R + R = \frac{9}{2}R$. Thus, $(b)$ matches with $(iv)$.
The correct matching is $(a-ii, b-iv)$.

(B) The heat capacity ratio $\gamma = \frac{C_{P}}{C_{V}}$ is a dimensionless quantity that provides information about the atomicity of the gas.
By knowing the value of $\gamma$,one can determine the degrees of freedom $(f)$ of the gas molecules using the relation $\gamma = 1 + \frac{2}{f}$.
This allows us to distinguish between monoatomic,diatomic,and polyatomic gases.

(N/A) The expression $C_P - C_V = R$ is Mayer's relation where $C_P$ and $C_V$ are molar specific heats measured in units of energy (Joules per mole per Kelvin) and $R$ is the universal gas constant $(8.314 \ J \ mol^{-1} K^{-1})$.
The expression $C_P - C_V = \frac{R}{J}$ is used when $C_P$ and $C_V$ are expressed in thermal units (calories per mole per Kelvin). Here,$J$ is the mechanical equivalent of heat $(J \approx 4.18 \ J/cal)$,which converts the energy unit $R$ into calories.
The expression $C_P - C_V = \frac{r}{J}$ is used when $C_P$ and $C_V$ are specific heats per unit mass (specific heat capacity) rather than molar specific heats. Here,$r$ is the specific gas constant,defined as $r = \frac{R}{M}$,where $M$ is the molar mass of the gas. This equation gives the difference in units of calories per gram per Kelvin.

(C) Given:
Heat at constant pressure: $Q_P = n C_P \Delta T_1 = 160 \text{ cal}$,where $\Delta T_1 = 50^{\circ} C$.
Heat at constant volume: $Q_V = n C_V \Delta T_2 = 240 \text{ cal}$,where $\Delta T_2 = 100^{\circ} C$.
From the first equation: $n C_P = \frac{160}{50} = 3.2$.
From the second equation: $n C_V = \frac{240}{100} = 2.4$.
Taking the ratio: $\frac{C_P}{C_V} = \gamma = \frac{3.2}{2.4} = \frac{4}{3}$.
The relation between degrees of freedom $f$ and adiabatic index $\gamma$ is $\gamma = 1 + \frac{2}{f}$.
Substituting $\gamma = \frac{4}{3}$: $\frac{4}{3} = 1 + \frac{2}{f} \Rightarrow \frac{1}{3} = \frac{2}{f} \Rightarrow f = 6$.

(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.
$(A)$ Monoatomic: $f = 3$,so $\gamma = 1 + \frac{2}{3} = \frac{5}{3}$ $(IV)$.
$(B)$ Diatomic rigid molecules: $f = 5$,so $\gamma = 1 + \frac{2}{5} = \frac{7}{5}$ $(I)$.
$(C)$ Diatomic non-rigid molecules: $f = 7$,so $\gamma = 1 + \frac{2}{7} = \frac{9}{7}$ $(II)$.
$(D)$ Triatomic rigid molecules: $f = 6$,so $\gamma = 1 + \frac{2}{6} = 1 + \frac{1}{3} = \frac{4}{3}$ $(III)$.
Thus,the correct matching is $A-IV, B-I, C-II, D-III$.

(C) According to the first law of thermodynamics, $\Delta Q = \Delta U + \Delta W$.
Since no work is done, $\Delta W = 0$.
Therefore, the heat supplied is equal to the change in internal energy: $\Delta Q = \Delta U = n C_v \Delta T$.
For a rigid diatomic gas, the molar heat capacity at constant volume is $C_v = \frac{5}{2} R$.
Given $n = 4 \, \text{moles}$ and $\Delta T = 50^{\circ} \text{C} - 0^{\circ} \text{C} = 50 \, \text{K}$.
Substituting the values: $\Delta Q = 4 \times \frac{5}{2} R \times 50 = 10 \times 50 \, R = 500 \, R$.

(B) The molar heat capacity at constant volume is given by $C_{V} = \frac{fR}{2}$,where $f$ is the degree of freedom and $R$ is the universal gas constant.
The molar heat capacity at constant pressure is given by Mayer's relation: $C_{P} = C_{V} + R$.
Substituting the value of $C_{V}$,we get $C_{P} = \frac{fR}{2} + R = R\left(1 + \frac{f}{2}\right) = R\left(\frac{f+2}{2}\right)$.
The ratio of specific heats $\gamma = \frac{C_{P}}{C_{V}}$ is calculated as:
$\gamma = \frac{R\left(\frac{f+2}{2}\right)}{\frac{fR}{2}} = \frac{f+2}{f} = 1 + \frac{2}{f}$.

(C) The internal energy $U$ of an ideal gas is given by the formula $U = n C_v T$.
For a monoatomic gas,the molar heat capacity at constant volume is $C_v = \frac{3}{2} R$.
Given: $n = 2\,mol$,$T = 300\,K$,and $R = 8.31\,J/mol\cdot K$.
Substituting these values into the formula:
$U = 2 \times \left( \frac{3}{2} R \right) \times 300$
$U = 3 \times R \times 300$
$U = 900 \times 8.31$
$U = 7479\,J$.

(C) The number of moles $n$ of the gas at $STP$ is given by the volume divided by the molar volume at $STP$ $(22.4 \, L/mol)$:
$n = \frac{44.8 \, L}{22.4 \, L/mol} = 2 \, mol$.
Since helium is a monoatomic gas,its molar heat capacity at constant volume is $C_V = \frac{3}{2} R$.
The heat required to raise the temperature by $\Delta T = 20.0^{\circ} C$ (which is equivalent to $20.0 \, K$) is given by the formula:
$\Delta Q = n C_V \Delta T$.
Substituting the values:
$\Delta Q = 2 \times (\frac{3}{2} R) \times 20.0$
$\Delta Q = 3 \times R \times 20.0 = 60 R$.
Given $R = 8.3 \, J \, K^{-1} \, mol^{-1}$:
$\Delta Q = 60 \times 8.3 = 498 \, J$.

(A) The molar specific heat at constant volume is given by $C_v = \frac{nR}{2}$.
The molar specific heat at constant pressure is given by $C_p = C_v + R = \frac{nR}{2} + R = \frac{(n+2)R}{2}$.
The ratio of specific heat at constant volume to the specific heat at constant pressure is $\frac{C_v}{C_p} = \frac{\frac{nR}{2}}{\frac{(n+2)R}{2}} = \frac{n}{n+2}$.

(B) The correct option is $(b)$.
Mayer's relation,$C_p - C_V = R$,is derived based on the equation of state for an ideal gas $(PV = nRT)$.
This relation holds true for any ideal gas,regardless of its atomicity (monoatomic,diatomic,or polyatomic).
For real gases,the relation is given by $C_p - C_V = TV \beta^2 / K_T$,where $K_T$ is the isothermal compressibility and $\beta$ is the isobaric thermal expansion coefficient. Since real gases do not follow the ideal gas law exactly,this specific relation $C_p - C_V = R$ is only strictly valid for ideal gases.

(D) The molar specific heat of a gas is defined as $C = \frac{dQ}{n dT}$.
For an adiabatic process (no heat exchange),$dQ = 0$,so $C_A = 0$.
For an isothermal process (constant temperature),$dT = 0$,so $C_T = \frac{dQ}{0} = \infty$.
For an isobaric process (constant pressure),$C_P = \frac{\gamma R}{\gamma - 1}$.
For an isochoric process (constant volume),$C_V = \frac{R}{\gamma - 1}$.
Comparing these with the given options,the statement $C_V = \frac{R}{\gamma}$ is incorrect because the correct formula is $C_V = \frac{R}{\gamma - 1}$.
Therefore,the correct option is $D$.

(B) For a monatomic gas,the adiabatic index is $\gamma = 5/3$.
At constant pressure,the heat supplied is given by $\Delta Q_P = n C_P \Delta T = 20 \, J$.
At constant volume,the heat supplied is given by $\Delta Q_V = n C_V \Delta T$.
We know that $C_P = \gamma C_V$,so $C_V = C_P / \gamma$.
Substituting this into the equation for constant volume:
$\Delta Q_V = n (C_P / \gamma) \Delta T = \frac{1}{\gamma} (n C_P \Delta T)$.
Substituting the given values:
$\Delta Q_V = \frac{1}{5/3} \times 20 = \frac{3}{5} \times 20 = 12 \, J$.

(C) The fraction of heat energy used to change the internal energy at constant pressure is given by the ratio $\frac{\Delta U}{\Delta Q}$.
Since $\Delta Q = n C_P \Delta T$ and $\Delta U = n C_V \Delta T$,we have:
$\frac{\Delta U}{\Delta Q} = \frac{n C_V \Delta T}{n C_P \Delta T} = \frac{C_V}{C_P} = \frac{1}{\gamma}$.
For a monatomic gas,$\gamma = \frac{5}{3}$,so $\frac{\Delta U}{\Delta Q} = \frac{1}{5/3} = \frac{3}{5} = 0.60$.
For a diatomic gas,$\gamma = \frac{7}{5}$,so $\frac{\Delta U}{\Delta Q} = \frac{1}{7/5} = \frac{5}{7} \approx 0.71$.
For a triatomic gas (non-linear),$\gamma = \frac{4}{3}$,so $\frac{\Delta U}{\Delta Q} = \frac{1}{4/3} = \frac{3}{4} = 0.75$.
Comparing the values,$0.75 > 0.71 > 0.60$. Thus,the fractional energy used to change internal energy is maximum in triatomic gases.

(B) At constant pressure, the heat absorbed is given by $\Delta Q_P = n C_P \Delta T$.
Given $\Delta Q_P = 105 \, cal$, $n = 3 \, moles$, and $\Delta T = 35^{\circ} C - 30^{\circ} C = 5^{\circ} C$.
At constant volume, the heat absorbed is given by $\Delta Q_V = n C_V \Delta T$.
We know that $\gamma = \frac{C_P}{C_V}$, which implies $C_P = \gamma C_V$.
Therefore, $\frac{\Delta Q_P}{\Delta Q_V} = \frac{n C_P \Delta T}{n C_V \Delta T} = \frac{C_P}{C_V} = \gamma$.
Substituting the values: $\frac{105}{\Delta Q_V} = 1.4$.
$\Delta Q_V = \frac{105}{1.4} = 75 \, cal$.
Since the temperature change $\Delta T$ is the same $(5^{\circ} C)$ in both cases, the heat required at constant volume is $75 \, cal$.

(A) For an ideal triatomic gas (non-linear),the degrees of freedom $f = 6$ (neglecting vibrational modes).
The molar heat capacity at constant pressure is given by $C_p = \frac{f+2}{2} R = \frac{6+2}{2} R = 4R$.
The heat supplied at constant pressure is $\Delta Q = n C_p \Delta T = n(4R) \Delta T = 800 \,cal$.
The work done by the gas at constant pressure is $\Delta W = n R \Delta T$.
From the heat equation,we have $n R \Delta T = \frac{800}{4} = 200 \,cal$.
Therefore,the energy used by the gas in work done against the surroundings is $\Delta W = 200 \,cal$.

(A) For a non-linear triatomic gas molecule,the degrees of freedom $(f)$ are calculated as follows:
$1$. Translational degrees of freedom: $3$ (along $x, y, z$ axes).
$2$. Rotational degrees of freedom: $3$ (since it is non-linear,it can rotate about three mutually perpendicular axes).
$3$. Total degrees of freedom: $f = 3 + 3 = 6$.
The molar specific heat at constant volume $(C_v)$ is given by the formula:
$C_v = \frac{f}{2} R$
Substituting $f = 6$ into the formula:
$C_v = \frac{6}{2} R = 3 R$
Therefore,the correct option is $A$.

(C) Given: Heat at constant pressure $Q_P = 50 \, cal$,number of moles $n = 1 \, mol$,change in temperature $\Delta T = 25^{\circ} C - 20^{\circ} C = 5 \, K$,and gas constant $R = 2 \, cal / mol \cdot K$.
First,we calculate the molar heat capacity at constant pressure $(C_P)$:
$Q_P = n C_P \Delta T$
$50 = 1 \times C_P \times 5$
$C_P = 10 \, cal / mol \cdot K$.
Using Mayer's relation,$C_P - C_V = R$,we find the molar heat capacity at constant volume $(C_V)$:
$C_V = C_P - R$
$C_V = 10 - 2 = 8 \, cal / mol \cdot K$.
Now,calculate the heat required at constant volume $(Q_V)$:
$Q_V = n C_V \Delta T$
$Q_V = 1 \times 8 \times 5 = 40 \, cal$.
Therefore,the correct option is $C$.