Calculate the kinetic energy of 5 mole of a gas at `27^(@)C` in erg and calorie

Answer

To calculate the kinetic energy of 5 moles of a gas at 27°C, we will follow these steps:

### Step 1: Convert the temperature from Celsius to Kelvin
To convert Celsius to Kelvin, we use the formula:
\[ T(K) = T(°C) + 273 \]

Given:
\[ T(°C) = 27 \]
\[ T(K) = 27 + 273 = 300 \, K \]

### Step 2: Use the kinetic energy formula
The kinetic energy (KE) of a gas can be calculated using the formula:
\[ KE = \frac{3}{2} nRT \]
where:
- \( n \) = number of moles of the gas
- \( R \) = universal gas constant
- \( T \) = temperature in Kelvin

### Step 3: Determine the value of R in calories
For calculations in calories, the value of \( R \) is:
\[ R = 2 \, \text{calories/(mole K)} \]

### Step 4: Substitute the values into the kinetic energy formula
Given:
- \( n = 5 \, \text{moles} \)
- \( R = 2 \, \text{calories/(mole K)} \)
- \( T = 300 \, K \)

Now substituting these values into the formula:
\[ KE = \frac{3}{2} \times 5 \times 2 \times 300 \]
\[ KE = \frac{3}{2} \times 5 \times 600 \]
\[ KE = \frac{3}{2} \times 3000 \]
\[ KE = 4500 \, \text{calories} \]

### Step 5: Convert the kinetic energy from calories to ergs
To convert calories to ergs, we use the conversion factor:
\[ 1 \, \text{calorie} = 418400 \, \text{ergs} \]

Now, converting:
\[ KE = 4500 \, \text{calories} \times 418400 \, \text{ergs/calorie} \]
\[ KE = 4500 \times 418400 \]
\[ KE = 1882800000 \, \text{ergs} \]

### Final Answer
The kinetic energy of 5 moles of gas at 27°C is:
- **In calories:** 4500 calories
- **In ergs:** 1882800000 ergs

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