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, we will use the ideal gas law and the relationship between pressure and volume. Let's break it down step by step.

### Step 1: Understand the Initial Conditions
- Bulb X has a volume \( V_X = 100 \, \text{cm}^3 \) and contains gas at pressure \( P \).
- Bulb Y is empty initially, so its volume \( V_Y \) is unknown.

### Step 2: Analyze the Change in Pressure
- When the stopcock is opened, the pressure in the system drops to \( 60\% \) of the original pressure \( P \).
- Therefore, the new pressure \( P' = 0.6P \).

### Step 3: Apply the Ideal Gas Law
Using the ideal gas law, we know that:
\[ P_1 V_1 = P_2 V_2 \]
Where:
- \( P_1 \) is the initial pressure in bulb X,
- \( V_1 \) is the initial volume of bulb X,
- \( P_2 \) is the new pressure after opening the stopcock,
- \( V_2 \) is the total volume of the system after opening the stopcock (which is \( V_X + V_Y \)).

### Step 4: Set Up the Equation
Substituting the known values into the equation:
\[ P \cdot 100 = 0.6P \cdot (100 + V_Y) \]

### Step 5: Simplify the Equation
We can cancel \( P \) from both sides (assuming \( P \neq 0 \)):
\[ 100 = 0.6(100 + V_Y) \]

### Step 6: Distribute and Rearrange
Distributing \( 0.6 \):
\[ 100 = 60 + 0.6V_Y \]

Now, subtract \( 60 \) from both sides:
\[ 40 = 0.6V_Y \]

### Step 7: Solve for \( V_Y \)
To find \( V_Y \), divide both sides by \( 0.6 \):
\[ V_Y = \frac{40}{0.6} = \frac{400}{6} \approx 66.67 \, \text{cm}^3 \]

### Final Answer
The volume of bulb Y must be approximately \( 66.67 \, \text{cm}^3 \).

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