Two glass bulbs x and y are connected by a very small tube having a stop - cock. Bulb X has a volume of `100cm^(3)` and contained the gas, while bulb Y was empty. On opening the stop - cock, the pressure fell down to `60%`. The volume of the bulb Y must be

Options

A. `132.32 cm^(3)`
B. `124cm^(3)`
C. `66.66cm^(3)`
D. `150cm^(3)`

Answer

To solve the problem step by step, we will use Boyle's Law, which states that the pressure of a gas multiplied by its volume is constant when the temperature is constant.

### Step 1: Define the initial conditions
Let:
- The initial pressure in bulb X be \( P \).
- The initial volume of bulb X be \( V_X = 100 \, \text{cm}^3 \).
- The volume of bulb Y be \( V_Y \) (which we need to find).

### Step 2: Determine the final pressure
When the stop-cock is opened, the pressure in the system falls to 60% of its initial value. Therefore, the final pressure \( P_f \) is:
\[ P_f = 0.6P = \frac{3}{5}P \]

### Step 3: Calculate the total volume after opening the stop-cock
After opening the stop-cock, the total volume \( V_T \) of the gas is the sum of the volumes of both bulbs:
\[ V_T = V_X + V_Y = 100 + V_Y \]

### Step 4: Apply Boyle's Law
According to Boyle's Law:
\[ P \cdot V_X = P_f \cdot V_T \]
Substituting the known values:
\[ P \cdot 100 = \left(\frac{3}{5}P\right) \cdot (100 + V_Y) \]

### Step 5: Simplify the equation
We can cancel \( P \) from both sides (assuming \( P \neq 0 \)):
\[ 100 = \frac{3}{5} (100 + V_Y) \]

### Step 6: Solve for \( V_Y \)
To eliminate the fraction, multiply both sides by 5:
\[ 500 = 3(100 + V_Y) \]
Expanding the right side:
\[ 500 = 300 + 3V_Y \]
Now, isolate \( V_Y \):
\[ 500 - 300 = 3V_Y \]
\[ 200 = 3V_Y \]
\[ V_Y = \frac{200}{3} \approx 66.67 \, \text{cm}^3 \]

### Step 7: Conclusion
The volume of bulb Y is approximately \( 66.67 \, \text{cm}^3 \).