The value of the ratio `C_(P)//C_(V)` for hydrogen is `1.67 at 300K` but decreases to `1.4 at 300k` as more degrees of freedom become active. During this rise in temperature (assume `H_(2)` is ideal gas).

Options

A. `C_(P)` remains constant but `C_(V)` increases
B. `C_(P)` decreases but `C_(V)` increases
C. Both `C_(P)` and `C_(V)` decrease by the same amount
D. Both `C_(P)` and `C_(V)` increase by the same amount

Answer

To solve the problem regarding the ratio of specific heats \( \frac{C_p}{C_v} \) for hydrogen gas as the temperature changes, we will follow these steps:

### Step 1: Understand the Given Information
We know that:
- At \( T_1 = 30 \, \text{K} \), \( \gamma_1 = \frac{C_p}{C_v} = 1.67 \)
- At \( T_2 = 300 \, \text{K} \), \( \gamma_2 = \frac{C_p}{C_v} = 1.4 \)

### Step 2: Use the Relationship Between \( C_v \) and \( C_p \)
The relationships for \( C_v \) and \( C_p \) are given by:
1. \( C_v = \frac{R}{\gamma - 1} \)
2. \( C_p = C_v + R \)

### Step 3: Calculate \( C_v \) and \( C_p \) at 30 K
Using the first formula for \( \gamma_1 \):
\[ C_{v1} = \frac{R}{\gamma_1 - 1} = \frac{R}{\frac{5}{3} - 1} = \frac{R}{\frac{2}{3}} = \frac{3R}{2} \]
Now, calculate \( C_{p1} \):
\[ C_{p1} = C_{v1} + R = \frac{3R}{2} + R = \frac{3R}{2} + \frac{2R}{2} = \frac{5R}{2} \]

### Step 4: Calculate \( C_v \) and \( C_p \) at 300 K
Using the second formula for \( \gamma_2 \):
\[ C_{v2} = \frac{R}{\gamma_2 - 1} = \frac{R}{\frac{7}{5} - 1} = \frac{R}{\frac{2}{5}} = \frac{5R}{2} \]
Now, calculate \( C_{p2} \):
\[ C_{p2} = C_{v2} + R = \frac{5R}{2} + R = \frac{5R}{2} + \frac{2R}{2} = \frac{7R}{2} \]

### Step 5: Analyze the Changes in \( C_v \) and \( C_p \)
- The change in \( C_v \) from \( 30 \, \text{K} \) to \( 300 \, \text{K} \):
\[ \Delta C_v = C_{v2} - C_{v1} = \frac{5R}{2} - \frac{3R}{2} = R \]
- The change in \( C_p \) from \( 30 \, \text{K} \) to \( 300 \, \text{K} \):
\[ \Delta C_p = C_{p2} - C_{p1} = \frac{7R}{2} - \frac{5R}{2} = R \]

### Step 6: Conclusion
Both \( C_p \) and \( C_v \) increase with temperature, and they increase by the same amount \( R \).