An amount of heat 50J of heat energy is removed from 4 moles of a monatomic ideal gas at constant volume.The temperature drops by

Options

A. `40^(@)C`
B. `30^(@)C`
C. `10^(@)C`
D. `0^(@)C`

Answer

To solve the problem of how much the temperature drops when 50 J of heat energy is removed from 4 moles of a monatomic ideal gas at constant volume, we can follow these steps:

### Step 1: Understand the First Law of Thermodynamics
The first law of thermodynamics states:
\[ \Delta Q = \Delta U + \Delta W \]
Where:
- \(\Delta Q\) is the heat added to the system,
- \(\Delta U\) is the change in internal energy,
- \(\Delta W\) is the work done by the system.

Since the process occurs at constant volume, the work done (\(\Delta W\)) is zero:
\[ \Delta W = 0 \]
Thus, we have:
\[ \Delta Q = \Delta U \]

### Step 2: Determine the Change in Internal Energy
Given that 50 J of heat is removed from the gas, we can express this as:
\[ \Delta Q = -50 \, \text{J} \]
Therefore:
\[ \Delta U = -50 \, \text{J} \]

### Step 3: Relate Change in Internal Energy to Temperature Change
For a monatomic ideal gas, the change in internal energy (\(\Delta U\)) can be expressed as:
\[ \Delta U = \frac{3}{2} n R \Delta T \]
Where:
- \(n\) is the number of moles,
- \(R\) is the universal gas constant (\(R \approx 8.314 \, \text{J/(mol K)}\)),
- \(\Delta T\) is the change in temperature.

### Step 4: Substitute Known Values
We know:
- \(n = 4 \, \text{moles}\),
- \(\Delta U = -50 \, \text{J}\).

Substituting these values into the equation:
\[ -50 = \frac{3}{2} \times 4 \times 8.314 \times \Delta T \]

### Step 5: Solve for \(\Delta T\)
First, calculate the constant:
\[ \frac{3}{2} \times 4 \times 8.314 = 49.884 \, \text{J/K} \]
Now, substituting this back into the equation:
\[ -50 = 49.884 \Delta T \]
To find \(\Delta T\):
\[ \Delta T = \frac{-50}{49.884} \approx -1.003 \, \text{K} \]

### Step 6: Convert to Degrees Celsius
Since a change of 1 K is equivalent to a change of 1 °C, we can say:
\[ \Delta T \approx -1.003 \, \text{°C} \]

### Final Answer
The temperature drops by approximately 1 °C.