An ideal gas is heated by `160J` at constant pressure, its temperature rises by `50^@` and if `240J` of heat is supplied at constant volume, temperature rises by `100^@C` , then its degree of freedom should be

Options

A. 3
B. 5
C. 6
D. 7

Answer

To solve the problem, we will follow these steps:

### Step 1: Understand the relationship between heat, temperature change, and specific heat at constant pressure (Cp).
When heat \( Q \) is added to an ideal gas at constant pressure, the relationship is given by:
\[ Q = n C_p \Delta T \]
where:
- \( Q \) is the heat added,
- \( n \) is the number of moles,
- \( C_p \) is the specific heat at constant pressure,
- \( \Delta T \) is the change in temperature.

### Step 2: Apply the equation for the first scenario (constant pressure).
From the problem, we know:
- \( Q = 160 \, J \)
- \( \Delta T = 50 \, °C \)

Substituting these values into the equation:
\[ 160 = n C_p (50) \]
This simplifies to:
\[ n C_p = \frac{160}{50} = 3.2 \quad \text{(1)} \]

### Step 3: Understand the relationship for heat at constant volume (Cv).
Similarly, when heat \( Q \) is added at constant volume:
\[ Q = n C_v \Delta T \]
where:
- \( C_v \) is the specific heat at constant volume.

### Step 4: Apply the equation for the second scenario (constant volume).
From the problem, we know:
- \( Q = 240 \, J \)
- \( \Delta T = 100 \, °C \)

Substituting these values into the equation:
\[ 240 = n C_v (100) \]
This simplifies to:
\[ n C_v = \frac{240}{100} = 2.4 \quad \text{(2)} \]

### Step 5: Relate Cp and Cv to the degree of freedom (F).
The relationship between \( C_p \) and \( C_v \) is given by:
\[ C_p - C_v = R \]
Also, we know:
\[ \gamma = \frac{C_p}{C_v} = 1 + \frac{2}{F} \]

### Step 6: Express \( C_p \) and \( C_v \) in terms of \( n \).
From equations (1) and (2):
\[ C_p = \frac{3.2}{n} \quad \text{and} \quad C_v = \frac{2.4}{n} \]

### Step 7: Substitute \( C_p \) and \( C_v \) into the equation for \( \gamma \).
Using the relationship:
\[ \gamma = \frac{C_p}{C_v} = \frac{\frac{3.2}{n}}{\frac{2.4}{n}} = \frac{3.2}{2.4} = \frac{32}{24} = \frac{4}{3} \]

### Step 8: Solve for \( F \).
Using the equation:
\[ \frac{4}{3} = 1 + \frac{2}{F} \]
Subtracting 1 from both sides:
\[ \frac{4}{3} - 1 = \frac{2}{F} \]
This simplifies to:
\[ \frac{1}{3} = \frac{2}{F} \]
Cross-multiplying gives:
\[ F = 6 \]

### Final Answer:
The degree of freedom \( F \) of the gas is **6**.

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