Question

If ${C_P}$ and ${C_V}$ denote the specific heats (per unit mass) of an ideal gas of molecular weight $M$, where $R$ is the molar gas constant.

(A) ${C_P} - {C_V} = R/{M^2}$

(B) ${C_P} - {C_V} = R$

(C) ${C_P} - {C_V} = R/M$

(D) ${C_P} - {C_V} = M/R$

Answer 1

Using the law of thermodynamics, we need to write the relations of ${C_P}$ and ${C_V}$. Substituting the quantities of these two relations with the quantities from laws of thermodynamics, we can derive a relation between ${C_P}$ and ${C_V}$.

Complete step by step solution:

We know the molar heat capacity of a substance at constant pressure is denoted by ${C_P}$ and heat transferred between a body and surrounding at constant volume is denoted by ${C_V}$. Or in other words, molar heat capacity of a substance at constant volume.

From First law of thermodynamics, we know if the increase in internal energy be denoted by $dU$and $dQ$ be the heat energy supplied, then work done $dW$ is denoted by:

$dQ = dU + dW$

Rewriting the equation of first law by introducing the concepts Entropy and enthalpy.

Entropy of a system is the degree of disorderness of a system.

Enthalpy is defined as the sum of internal energy and product of pressure and volume.

Therefore, enthalpy is defined as:

$H = U + PV$

Where,

$H$ is the enthalpy and $U$is the internal energy.

Thus, we can write:

$dH = dU + d(PV)$

We know, that

$PV = nRT$

Putting this values, in the above equation, we get:

$dH = dU + nRdT$

We also know, from the equation of specific heat, that:

$dH = {mC_P}dT$

Therefore, the equation becomes:

${mC_P}dT = {mC_v}dT + nRdT$

Cancelling the common term from both sides,

${mC_P} = {mC_v} + nR$

$\Rightarrow {mC_P} - {mC_v} = nR$

This can be written as:

${mC_P} - {mC_v} = nR = \dfrac{{Rm}}{M}$

Thus, we obtain ${C_P} - {C_v} = \dfrac{R}{M}$

This is our required solution.

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Hence, option Cis correct.

Note: The value of ${C_P}$ is always greater than${C_v}$. This happens because at constant pressure, when heat is added to the substance, expansion occurs in the substance and hence work is done.