When 1 mole of an ideal monoatomic gas is compressed adiabatically the internal energy change involved is 24 cals. The temperature rise (in kelwin) is

Answer

To solve the problem step by step, we will follow the principles of thermodynamics related to ideal gases, particularly focusing on the internal energy change during an adiabatic process.

### Step-by-Step Solution:

1. **Understanding the Given Information:**
We know that:
- The internal energy change (ΔU) is given as 24 calories.
- We have 1 mole of an ideal monoatomic gas.

2. **Using the Formula for Internal Energy Change:**
The change in internal energy (ΔU) for an ideal gas can be expressed as:
\[ \Delta U = n C_v \Delta T \]
where:
- \( n \) is the number of moles,
- \( C_v \) is the molar heat capacity at constant volume,
- \( \Delta T \) is the change in temperature.

3. **Identifying the Value of \( C_v \) for a Monoatomic Gas:**
For a monoatomic ideal gas, the molar heat capacity at constant volume \( C_v \) is:
\[ C_v = \frac{3}{2} R \]
where \( R \) is the universal gas constant.

4. **Substituting Values:**
We need to use the value of \( R \) in calories. The value of \( R \) in calories per mole per Kelvin is:
\[ R = 2 \text{ cal/(mol K)} \]
Now substituting \( n = 1 \) mole and \( C_v = \frac{3}{2} R \):
\[ C_v = \frac{3}{2} \times 2 = 3 \text{ cal/(mol K)} \]

5. **Setting Up the Equation:**
Now, substituting the known values into the internal energy change equation:
\[ 24 \text{ cal} = 1 \text{ mol} \times 3 \text{ cal/(mol K)} \times \Delta T \]

6. **Solving for \( \Delta T \):**
Rearranging the equation to solve for \( \Delta T \):
\[ \Delta T = \frac{24 \text{ cal}}{3 \text{ cal/(mol K)}} \]
\[ \Delta T = 8 \text{ K} \]

### Final Answer:
The temperature rise (ΔT) is **8 Kelvin**.

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