
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 molecular weight constant, then
(A) ${C_p} - {C_v} = \dfrac{R}{{{M^2}}}$
(B) ${C_p} - {C_v} = R$
(C) ${C_p} - {C_v} = \dfrac{R}{M}$
(D) ${C_p} - {C_v} = MR$
Answer
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Hint: The difference of the ideal gas at constant pressure and the constant volume can be determined by using the specific heat formula of the ideal gas equation. And then by using Mayer’s formula in thermodynamics, the solution can be determined.
Formula used:
Mayer’s formula in thermodynamics is given by,
${C_p} - {C_v} = R$
Where, ${C_p}$ is the molar specific heat of the ideal gas at constant pressure, ${C_v}$ is the molar specific heat of the ideal gas at constant pressure and $R$ is the molecular weight constant of the ideal gas.
Complete step by step answer:
Given that,
The molecular weight of the ideal gas is, $M$.
The molecular weight constant of the ideal gas is, $R$.
Mayer’s formula in thermodynamics is given by,
${C_p} - {C_v} = R\,..................\left( 1 \right)$
Now,
Let ${C_v}$ and ${C_p}$ be the molar specific heats of the ideal gas at constant volume and constant pressure, then the molecular weight of the ideal gas is $M$, then
For ideal gas at constant pressure is given by,
$\Rightarrow {C_p} = M \times {C_p}\,................\left( 2 \right)$
For ideal gas at constant volume is given by,
$\Rightarrow {C_v} = M \times {C_v}\,................\left( 3 \right)$
By substituting the equation (2) and equation (3) in the equation (1), then the equation (1) is written as,
$\Rightarrow M{C_p} - M{C_v} = R$
By taking the term molecular weight of an ideal gas $M$ as common, then the above equation is written as,
$\Rightarrow M\left( {{C_p} - {C_v}} \right) = R$
By taking the term molecular weight of an ideal gas $M$ from LHS to RHS, then the above equation is written as,
$\Rightarrow {C_p} - {C_v} = \dfrac{R}{M}$
Thus, the above equation shows the specific heats per unit mass.
Hence, option (C) is the correct answer.
Note:
The specific heat of an ideal gas at constant pressure ${C_p}$ and the specific heat of an ideal gas at constant volume ${C_v}$ are also used in the first law of thermodynamics equation. Also, specific heat is used in the heat capacity equation.
Formula used:
Mayer’s formula in thermodynamics is given by,
${C_p} - {C_v} = R$
Where, ${C_p}$ is the molar specific heat of the ideal gas at constant pressure, ${C_v}$ is the molar specific heat of the ideal gas at constant pressure and $R$ is the molecular weight constant of the ideal gas.
Complete step by step answer:
Given that,
The molecular weight of the ideal gas is, $M$.
The molecular weight constant of the ideal gas is, $R$.
Mayer’s formula in thermodynamics is given by,
${C_p} - {C_v} = R\,..................\left( 1 \right)$
Now,
Let ${C_v}$ and ${C_p}$ be the molar specific heats of the ideal gas at constant volume and constant pressure, then the molecular weight of the ideal gas is $M$, then
For ideal gas at constant pressure is given by,
$\Rightarrow {C_p} = M \times {C_p}\,................\left( 2 \right)$
For ideal gas at constant volume is given by,
$\Rightarrow {C_v} = M \times {C_v}\,................\left( 3 \right)$
By substituting the equation (2) and equation (3) in the equation (1), then the equation (1) is written as,
$\Rightarrow M{C_p} - M{C_v} = R$
By taking the term molecular weight of an ideal gas $M$ as common, then the above equation is written as,
$\Rightarrow M\left( {{C_p} - {C_v}} \right) = R$
By taking the term molecular weight of an ideal gas $M$ from LHS to RHS, then the above equation is written as,
$\Rightarrow {C_p} - {C_v} = \dfrac{R}{M}$
Thus, the above equation shows the specific heats per unit mass.
Hence, option (C) is the correct answer.
Note:
The specific heat of an ideal gas at constant pressure ${C_p}$ and the specific heat of an ideal gas at constant volume ${C_v}$ are also used in the first law of thermodynamics equation. Also, specific heat is used in the heat capacity equation.
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