Question

The internal energy of air in $ 4 \, \text{m} \times 4 \, \text{m} \times 3 \, \text{m} $ sized room at 1 atmospheric pressure will be $ \times 10^6 \, \text{J} $. (Consider air as a diatomic molecule)

Hint:

For a diatomic molecule, the internal energy is calculated using the formula \( U = \frac{5}{2} P V \), where \( P \) is pressure and \( V \) is volume.

Answer 1

Correct Answer: 12

To find the internal energy of gas in the room. \[ U = n C_v T = \frac{5}{2} R T \] \[ = \frac{5}{2} \times P V = \frac{5}{2} \times 10^5 \times 48 = 12 \times 10^6 \, \text{J} \]

Answer 2

We are given a room of size \(4\,\text{m} \times 4\,\text{m} \times 3\,\text{m}\), filled with air at 1 atm pressure. We must find its total internal energy assuming air behaves as a diatomic ideal gas.

Concept Used:

For an ideal gas, internal energy is given by:

\[ U = nC_vT \]

where \(n\) is the number of moles and \(C_v\) is the molar specific heat at constant volume.

For a diatomic gas (air), \(C_v = \frac{5}{2}R.\)

Step-by-Step Solution:

Step 1: Calculate the volume of the room.

\[ V = 4\times4\times3 = 48\,\text{m}^3. \]

Step 2: Using the ideal gas equation \( PV = nRT \), we can express the number of moles as:

\[ n = \frac{PV}{RT}. \]

Step 3: Internal energy per mole of gas is \( U = nC_vT = n \frac{5}{2}RT \).

Substitute \( n = \frac{PV}{RT} \):

\[ U = \frac{PV}{RT} \cdot \frac{5}{2}RT = \frac{5}{2}PV. \]

Step 4: Substitute the values:

\[ P = 1\,\text{atm} = 1.013\times10^5\,\text{Pa}, \quad V = 48\,\text{m}^3. \] \[ U = \frac{5}{2}\times(1.013\times10^5)\times48. \] \[ U = 2.5 \times 1.013\times10^5 \times 48 = 1.2156\times10^7\,\text{J}. \]

Final Computation & Result:

\[ U = 12.16\times10^6\,\text{J} \approx 12.2\times10^6\,\text{J}. \]

Answer: \( \boxed{12.2\times10^6\ \text{J}} \)