Question

A glass bulb of volume 400 cm\(^3\) is connected to another bulb of volume 200 cm\(^3\) by means of a tube of negligible volume. The bulbs contain dry air. They are both at a common temperature and pressure of 20°C and 1.000 atm. The larger bulb is immersed in steam at 100°C; the smaller, in melting ice at 0°C. The final common pressure is:

• A. 2.31 atm
• B. 1.13 atm
• C. 0.53 atm
• D. 0.04 atm

Hint:

Use the combined gas law to find the final pressure when the temperature changes for a given volume.

Answer 1

The Correct Option is B

Using the combined gas law: \[ \frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2} \] where:
- \( P_1 = 1 \, \text{atm} \),
- \( V_1 = 400 + 200 = 600 \, \text{cm}^3 \),
- \( T_1 = 20°C = 293 \, \text{K} \),
- \( P_2 \) is the final pressure,
- \( V_2 = 600 \, \text{cm}^3 \) (since volume is constant),
- \( T_2 = 100°C = 373 \, \text{K} \). Substituting the values into the combined gas law: \[ \frac{1 \times 600}{293} = \frac{P_2 \times 600}{373} \] Solving for \( P_2 \): \[ P_2 = \frac{1 \times 373}{293} \approx 1.13 \, \text{atm} \] Thus, the final pressure is 1.13 atm.