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Step-by-Step Solution
Step 1: Identify the Given Data
• Initial load the balloon carries: 185 kg
• Initial barometric pressure, $P_{1}$: 76 cm of Hg (normal atmospheric pressure)
• Initial temperature, $T_{1}$: 27°C (which is 300 K)
• Final barometric pressure, $P_{2}$: 45 cm of Hg
• Final temperature, $T_{2}$: –7°C (which is 266 K)
• Balloon volume is assumed constant.
Step 2: Note the Relation Between Density, Pressure, and Temperature
For an ideal gas at constant volume,
$ P = \rho R T \quad \Longrightarrow \quad \rho \propto \frac{P}{T} \,. $
Therefore,
$ \frac{\rho_{1}}{\rho_{2}} = \frac{P_{1} \, / \, T_{1}}{P_{2} \, / \, T_{2}} = \frac{P_{1} \, T_{2}}{P_{2} \, T_{1}} \,. $
Step 3: Relate Density to the Load Carried
The load a balloon can carry is inversely proportional to the density of the surrounding air (or equivalently, directly proportional to the density difference if it is helium or hot air). Here, we compare loads by the ratio of densities:
$ \frac{M_{1}}{M_{2}} = \frac{\rho_{1}}{\rho_{2}} \,. $
Hence,
$ M_{2} = M_{1} \times \frac{\rho_{2}}{\rho_{1}} = M_{1} \times \frac{P_{2} \, T_{1}}{P_{1} \, T_{2}} \,. $
Step 4: Substitute the Numerical Values
Substitute the known values into the expression:
$ M_{2} = 185 \times \frac{(45 \text{ cm Hg}) \,(300 \text{ K})}{(76 \text{ cm Hg}) \,(266 \text{ K})} \,. $
Step 5: Calculate the Final Load
Perform the calculation:
$ M_{2} = 185 \times \frac{45 \times 300}{76 \times 266} \approx 123.54 \text{ kg} \,. $
Thus, at the reduced pressure and lower temperature (while keeping the balloon’s volume constant), the balloon can carry about 123.54 kg.
Final Answer
The load the balloon will carry is 123.54 kg.
