A bulb of unknown volume V contains a gas at 1 atm pressure. This bulb was connected to another evacuated bulb of volume 0.5 L through a stop cock. When the stop cock was opened, the pressure at each bulb becomes `7.58xx10^(4)"Pa mm"` while the temperature remained constant. Calculate V in litres.

Explanation

To solve the problem, we will use the ideal gas law and the principle of conservation of moles. Here’s a step-by-step solution:

### Step 1: Understand the Initial Conditions
We have two bulbs:
- Bulb 1 (unknown volume V) containing gas at an initial pressure of 1 atm.
- Bulb 2 (known volume 0.5 L) is evacuated (pressure = 0 atm).

### Step 2: Convert Pressure Units
The final pressure after opening the stopcock is given as \( 7.58 \times 10^4 \) Pa. We need to convert this pressure into atm for consistency.
- 1 atm = 101325 Pa
- Therefore, \( P_{final} = \frac{7.58 \times 10^4 \text{ Pa}}{101325 \text{ Pa/atm}} \approx 0.748 \text{ atm} \).

### Step 3: Apply the Ideal Gas Law
Using the ideal gas law, we know that:
\[ PV = nRT \]
Since the temperature (T) and the gas constant (R) remain constant, we can equate the initial and final states of the gas.

### Step 4: Set Up the Equation
Before opening the stopcock:
- For Bulb 1:
- \( P_1 = 1 \text{ atm} \)
- \( V_1 = V \)
- For Bulb 2:
- \( P_2 = 0 \text{ atm} \)
- \( V_2 = 0.5 \text{ L} \)

After opening the stopcock, the total pressure and volume become:
- \( P_{final} = 0.748 \text{ atm} \)
- \( V_{final} = V + 0.5 \text{ L} \)

Using the conservation of moles:
\[ P_1V_1 + P_2V_2 = P_{final}V_{final} \]
Substituting the known values:
\[ (1 \text{ atm})(V) + (0 \text{ atm})(0.5 \text{ L}) = (0.748 \text{ atm})(V + 0.5 \text{ L}) \]

### Step 5: Solve for V
This simplifies to:
\[ V = 0.748(V + 0.5) \]
Expanding the right side:
\[ V = 0.748V + 0.374 \]
Rearranging gives:
\[ V - 0.748V = 0.374 \]
\[ 0.252V = 0.374 \]
Now, solving for \( V \):
\[ V = \frac{0.374}{0.252} \approx 1.48 \text{ L} \]

### Final Answer
The volume \( V \) of the first bulb is approximately **1.48 liters**.

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