At a temperature `T, K`, the pressure of `4.0 gm` argon in a bulb is `P`. The bulb is put in a bath having temperature higher by `50 K` than the first one `0.8` of argon gas had to be removed to maintain original pressure. The temperature `T` is

Options

A. 510K
B. 200K
C. 100K
D. 73K

Answer

To solve the problem, we will use the ideal gas law and the relationship between the number of moles, temperature, and pressure. Here’s the step-by-step solution:

### Step 1: Understand the Ideal Gas Law
The ideal gas law is given by the equation:
\[ PV = nRT \]
Where:
- \( P \) = Pressure
- \( V \) = Volume
- \( n \) = Number of moles
- \( R \) = Ideal gas constant
- \( T \) = Temperature in Kelvin

Since the pressure and volume are constant, we can relate the number of moles and temperature.

### Step 2: Establish the Relationship
If pressure \( P \) and volume \( V \) are constant, we can say:
\[ n_1 T_1 = n_2 T_2 \]
Where:
- \( n_1 \) and \( T_1 \) are the initial number of moles and temperature.
- \( n_2 \) and \( T_2 \) are the final number of moles and temperature.

### Step 3: Calculate Initial Moles
The initial mass of argon is given as 4.0 g. The molar mass of argon (Ar) is approximately 40 g/mol. Therefore, the initial number of moles \( n_1 \) can be calculated as:
\[ n_1 = \frac{m_1}{M} = \frac{4.0 \, \text{g}}{40 \, \text{g/mol}} = 0.1 \, \text{mol} \]

### Step 4: Determine Final Moles
After removing 0.8 g of argon, the final mass \( m_2 \) is:
\[ m_2 = 4.0 \, \text{g} - 0.8 \, \text{g} = 3.2 \, \text{g} \]
Now, calculate the final number of moles \( n_2 \):
\[ n_2 = \frac{m_2}{M} = \frac{3.2 \, \text{g}}{40 \, \text{g/mol}} = 0.08 \, \text{mol} \]

### Step 5: Set Up the Equation
Now we can set up the equation using the relationship established earlier:
\[ n_1 T_1 = n_2 T_2 \]
Substituting the known values:
\[ (0.1 \, \text{mol}) (T) = (0.08 \, \text{mol}) (T + 50) \]

### Step 6: Solve for Temperature \( T \)
Expanding the equation gives:
\[ 0.1T = 0.08T + 4 \]
Now, isolate \( T \):
\[ 0.1T - 0.08T = 4 \]
\[ 0.02T = 4 \]
\[ T = \frac{4}{0.02} = 200 \, \text{K} \]

### Final Answer
The temperature \( T \) is:
\[ T = 200 \, \text{K} \]

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