In the temperature of 1 mole of a gas is increased by `50^(@)C`. Calculate chagne in kinetic energy:

Options

A. 62.32 J
B. 6.235J
C. 623.5J
D. 6325.0J

Explanation

To calculate the change in kinetic energy when the temperature of 1 mole of gas is increased by \(50^\circ C\), we can use the formula for the kinetic energy of an ideal gas:

\[ KE = \frac{3}{2} nRT \]

Where:
- \(KE\) is the kinetic energy,
- \(n\) is the number of moles,
- \(R\) is the universal gas constant (\(R \approx 8.314 \, \text{J/(mol K)}\)),
- \(T\) is the absolute temperature in Kelvin.

### Step 1: Identify the change in temperature
The change in temperature (\(\Delta T\)) is given as \(50^\circ C\). In Kelvin, this change is the same:

\[ \Delta T = 50 \, \text{K} \]

### Step 2: Calculate the change in kinetic energy
Since we are interested in the change in kinetic energy, we can express it as:

\[ \Delta KE = \frac{3}{2} nR \Delta T \]

Substituting the values:
- \(n = 1\) mole,
- \(R = 8.314 \, \text{J/(mol K)}\),
- \(\Delta T = 50 \, \text{K}\),

We get:

\[ \Delta KE = \frac{3}{2} \times 1 \times 8.314 \times 50 \]

### Step 3: Perform the calculation
Calculating the above expression:

\[ \Delta KE = \frac{3}{2} \times 8.314 \times 50 \]
\[ = 1.5 \times 8.314 \times 50 \]
\[ = 1.5 \times 415.7 \]
\[ = 623.55 \, \text{J} \]

### Final Answer
The change in kinetic energy when the temperature of 1 mole of gas is increased by \(50^\circ C\) is approximately:

\[ \Delta KE \approx 623.5 \, \text{J} \]

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