Two flask A and B of equal volumes maintained at temperature `300K` and `700K` contain equal mass of `He(g)` and `N_(2)(g)` respectively. What is the ratio of total translational kinetic energy of gas in flask A to that of flask B ?

Options

A. `1:3`
B. `3:1`
C. `3:49`
D. None of these

Explanation

To find the ratio of total translational kinetic energy of gas in flask A (containing He) to that of flask B (containing N₂), we can follow these steps:

### Step 1: Understand the formula for translational kinetic energy
The translational kinetic energy (KE) of an ideal gas can be expressed as:
\[ KE = \frac{3}{2} nRT \]
where \( n \) is the number of moles, \( R \) is the universal gas constant, and \( T \) is the temperature in Kelvin.

### Step 2: Determine the number of moles of each gas
Let the mass of helium in flask A be \( m \) grams. The molar mass of helium (He) is approximately \( 4 \, \text{g/mol} \).
Thus, the number of moles of He in flask A is:
\[ n_A = \frac{m}{4} \]

For flask B, which contains nitrogen (N₂), the molar mass of nitrogen is approximately \( 28 \, \text{g/mol} \).
Thus, the number of moles of N₂ in flask B is:
\[ n_B = \frac{m}{28} \]

### Step 3: Calculate the translational kinetic energy for both flasks
Now we can calculate the translational kinetic energy for both gases.

For flask A (He at \( 300 \, K \)):
\[ KE_A = \frac{3}{2} n_A R T_A = \frac{3}{2} \left(\frac{m}{4}\right) R (300) \]
\[ KE_A = \frac{3mR \cdot 300}{8} \]

For flask B (N₂ at \( 700 \, K \)):
\[ KE_B = \frac{3}{2} n_B R T_B = \frac{3}{2} \left(\frac{m}{28}\right) R (700) \]
\[ KE_B = \frac{3mR \cdot 700}{56} \]

### Step 4: Simplify the expressions for kinetic energy
Now we can simplify both expressions:
\[ KE_A = \frac{3mR \cdot 300}{8} = \frac{900mR}{8} = \frac{225mR}{2} \]
\[ KE_B = \frac{3mR \cdot 700}{56} = \frac{2100mR}{56} = \frac{525mR}{14} \]

### Step 5: Find the ratio of kinetic energies
Now we can find the ratio of \( KE_A \) to \( KE_B \):
\[ \text{Ratio} = \frac{KE_A}{KE_B} = \frac{\frac{225mR}{2}}{\frac{525mR}{14}} = \frac{225 \cdot 14}{525 \cdot 2} \]
\[ = \frac{3150}{1050} = 3 \]

### Final Answer
The ratio of total translational kinetic energy of gas in flask A to that of flask B is:
\[ \text{Ratio} = 3:1 \]