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, we need to determine the temperature rise (ΔT) when 1 mole of an ideal monoatomic gas is compressed adiabatically, given that the change in internal energy (ΔU) is 24 calories.

### Step-by-Step Solution:

1. **Understand the relationship between internal energy change and temperature change**:
The change in internal energy (ΔU) for an ideal gas can be expressed as:
\[ \Delta U = n C_V \Delta T \]
where:
- \( n \) = number of moles of the gas
- \( C_V \) = molar heat capacity at constant volume
- \( \Delta T \) = change in temperature

2. **Identify the values**:
- Given: \( \Delta U = 24 \) calories
- Number of moles \( n = 1 \) mole
- 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. In calories, \( R = 2 \) cal/(mol·K).

3. **Substitute the values into the equation**:
First, calculate \( C_V \):
\[ C_V = \frac{3}{2} \times 2 \text{ cal/(mol·K)} = 3 \text{ cal/(mol·K)} \]
Now, substitute \( n \), \( C_V \), and \( \Delta U \) into the internal energy equation:
\[ 24 \text{ cal} = 1 \text{ mol} \times 3 \text{ cal/(mol·K)} \times \Delta T \]

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

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

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