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 use the concepts of heat transfer in an ideal gas at constant pressure and constant volume, along with the relationship between molar specific heats.

### Step-by-Step Solution:

1. **Identify the Given Information:**
- Heat added at constant pressure, \( Q_P = 160 \, J \)
- Temperature rise at constant pressure, \( \Delta T_P = 50^\circ C \)
- Heat added at constant volume, \( Q_V = 240 \, J \)
- Temperature rise at constant volume, \( \Delta T_V = 100^\circ C \)

2. **Use the Formula for Heat Transfer:**
- For constant pressure:
\[ Q_P = n C_P \Delta T_P \]
- For constant volume:
\[ Q_V = n C_V \Delta T_V \]

3. **Set Up the Equations:**
- From the first equation (constant pressure):
\[ 160 = n C_P (50) \]
Rearranging gives:
\[ n C_P = \frac{160}{50} = 3.2 \, n \]
- From the second equation (constant volume):
\[ 240 = n C_V (100) \]
Rearranging gives:
\[ n C_V = \frac{240}{100} = 2.4 \, n \]

4. **Divide the Two Equations:**
- Dividing the two equations:
\[ \frac{n C_P}{n C_V} = \frac{3.2}{2.4} \]
- Simplifying gives:
\[ \frac{C_P}{C_V} = \frac{3.2}{2.4} = \frac{4}{3} \]

5. **Use the Relationship Between \( C_P \) and \( C_V \):**
- We know that:
\[ C_P = C_V + R \]
- Substitute \( C_P \) in terms of \( C_V \):
\[ \frac{C_V + R}{C_V} = \frac{4}{3} \]
- Cross-multiplying gives:
\[ 3(C_V + R) = 4C_V \]
- Rearranging yields:
\[ 3R = 4C_V - 3C_V \implies 3R = C_V \]

6. **Relate \( C_V \) to Degrees of Freedom:**
- The molar specific heat at constant volume is given by:
\[ C_V = \frac{F}{2} R \]
- Substituting for \( C_V \):
\[ 3R = \frac{F}{2} R \]
- Cancel \( R \) (assuming \( R \neq 0 \)):
\[ 3 = \frac{F}{2} \]
- Solving for \( F \):
\[ F = 6 \]

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