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

To solve this problem, we apply the Ideal Gas Law, given by the equation \(PV = nRT\), where \(P\) is pressure, \(V\) is volume, \(n\) is the amount of substance in moles, \(R\) is the gas constant, and \(T\) is the temperature in Kelvin. Initially, the two bulbs (400 cm\(^3\) and 200 cm\(^3\)) are at 20°C, and 1.000 atm. Converting temperatures to Kelvin, we have:

\(T_1 = 20°C = 293.15\ K\)

When the larger bulb is placed in steam, its temperature becomes: 

\(T_{L} = 100°C = 373.15\ K\)

The smaller bulb in ice has a temperature:

\(T_{S} = 0°C = 273.15\ K\)

Initially, the total volume is 600 cm\(^3\) with air at 293.15 K and 1.000 atm pressure. Using the Ideal Gas Law for the whole system:

\(P_iV_i = nRT_i\), where \(P_i = 1.000\ atm\), \(V_i = 0.6\ L\), \(n\) and \(R\) remain constant for the system.

After the temperature change when the larger bulb is at 373.15 K and the smaller at 273.15 K, and the pressure equilibrates to \(P_f\), the equation for the larger bulb is:

\(P_f \cdot 0.4\ L = n_L \cdot R \cdot 373.15\ K\)

And the equation for the smaller bulb is:

\(P_f \cdot 0.2\ L = n_S \cdot R \cdot 273.15\ K\)

Since air is distributed according to temperatures, \(n_i\) is conserved:

\(n_L = \frac{0.4}{0.6}n_i\)

\(n_S = \frac{0.2}{0.6}n_i\)

Using the Ideal Gas Law for both bulbs and combining terms, we get:

\((P_f \cdot 0.4 \cdot 273.15) + (P_f \cdot 0.2 \cdot 373.15) = (1.000 \cdot 0.6 \cdot 293.15)\)

Combining,

\(P_f \cdot (0.4 \cdot 273.15 + 0.2 \cdot 373.15) = (0.6 \cdot 293.15)\)

Simplifying gives:

\(P_f \cdot (109.26 + 74.63) = 175.89\)

\(P_f \cdot 183.89 = 175.89\)

Solving for \(P_f\),

\(P_f = \frac{175.89}{183.89} ≈ 1.13\ atm\)

Thus, the final common pressure is 1.13 atm.

Answer 2

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.