Question

The ratio \[{{C}_{p}}/{{C}_{v}}\]for ${{H}_{2}}$ is:
A. 1.40
B. 1.67
C. 1.33
D. None of these

Answer 1

Hint: \[{{C}_{p}}\] is the molar heat capacity of the molecule at constant pressure and ${{C}_{v}}$ is the molar heat capacity at constant volume. Identify the relationship between the heat capacity at constant volume and pressure. Find a diatomic gas's heat capacity at constant pressure. Using this value as a reference point for the relationship between the heat capacity at constant pressure and volume. Find the ratio of the two to get the required answer.

Formula Used: \[{{C}_{p}}={{C}_{v}}+R\], where \[{{C}_{p}}\]is the molar heat capacity of the molecule at constant pressure and ${{C}_{v}}$ is the molar heat capacity at constant volume and R is the universal gas constant

Complete Step by Step Answer:
It is possible to define the molar heat capacity of a material or substance as the amount of energy needed to raise the temperature of one mole of the substance by one degree, or one unit. \[{{C}_{p}}\]is the molar heat capacity of the molecule at constant pressure and ${{C}_{v}}$ is the molar heat capacity at constant volume. Since ${{H}_{2}}$ is a diatomic molecule we’ll find the value of \[{{C}_{p}}/{{C}_{v}}\] according to that. We have a diatomic gas,
The ${{C}_{v}}$ for an ideal gas is given by ${{C}_{v}}=\dfrac{5}{2}R$, where R is the universal gas constant.

We will make use of the formula given below:
\[{{C}_{p}}={{C}_{v}}+R\]
Therefore, ${{C}_{p}}=\dfrac{5}{2}R+R$
And hence we get ${{C}_{p}}=\dfrac{7}{2}R$

Now, in the question we have been asked to find the ratio of \[{{C}_{p}}/{{C}_{v}}\]. So we will divide the values of \[{{C}_{p}}\]and ${{C}_{v}}$
\[\dfrac{{{C}_{p}}}{{{C}_{v}}}=\dfrac{\dfrac{7}{2}R}{\dfrac{5}{2}R}=\dfrac{7}{5}=1.40\]
Hence, the correct option is A. 1.40.



Note: The specific heat capacity will vary for various gases. The specific heat capacity at constant volume of a monatomic gas is $\dfrac{3}{2}R$. The specific heat capacity at constant volume of a polyatomic gas is 3R.