A balloon carries a total load of 185 kg at normal pressure and temperature of 27°C. What load will the balloon carry on rising to a height at which the barometric pressure is 45 cm of Hg and the temperature is -7°C. Assuming the volume constant?

We begin with the principle of buoyancy for a gas balloon. The upward (lifting) force equals the weight of the air displaced, while the downward forces are the weight of the gas inside the balloon plus the weight of the load that the balloon can carry. If the material of the balloon itself is light enough to be neglected, the net load that can be supported is proportional to the difference between the density of the outside air and the density of the gas inside the balloon, multiplied by the volume of the balloon and the acceleration due to gravity.

Mathematically, for any set of atmospheric conditions, we may write

$$\text{Load}=Vg(\rho_{\text{air}}-\rho_{\text{gas}}).$$

Because the problem states “assuming the volume constant,” the volume $$V$$ (and of course $$g$$) do not change when the balloon rises. Hence the load varies only with the factor $$(\rho_{\text{air}}-\rho_{\text{gas}})$$.

Both the outside air and the gas inside the balloon obey the ideal-gas equation $$\rho=\frac{PM}{RT},$$ where $$P$$ is the pressure, $$T$$ is the absolute temperature, $$M$$ is the molar mass of the gas concerned and $$R$$ is the universal gas constant.

Inside the balloon the gas (say hydrogen or helium) is at the same pressure and temperature as the surrounding air, so the factor $$\dfrac{P}{T}$$ is common to both $$\rho_{\text{air}}$$ and $$\rho_{\text{gas}}$$. Therefore the difference of the two densities can be expressed as

$$\rho_{\text{air}}-\rho_{\text{gas}}=\frac{P}{RT}\bigl(M_{\text{air}}-M_{\text{gas}}\bigr).$$

Notice that the bracketed term $$(M_{\text{air}}-M_{\text{gas}})$$ is a constant, because the molar masses of air ($$\approx 29\ \text{g mol}^{-1}$$) and of the balloon gas ($$\approx 2\ \text{g mol}^{-1}$$ for hydrogen or $$4\ \text{g mol}^{-1}$$ for helium) do not change with height. Consequently, the entire density difference, and hence the load, varies directly as the ratio $$\dfrac{P}{T}$$.

So, for two different levels (ground and high-altitude) we have the proportionality

$$\frac{\text{Load at height}}{\text{Load at ground}}=\frac{\dfrac{P_2}{T_2}}{\dfrac{P_1}{T_1}}.$$

All the data are now substituted step by step. The ground (initial) conditions are stated to be “normal pressure” and $$27^{\circ}\text{C}$$. Normal pressure is $$P_1=76\ \text{cm of Hg},$$ and the absolute temperature is $$T_1=27^{\circ}\text{C}=27+273=300\ \text{K}.$$

At the new height the given pressure and temperature are

$$P_2=45\ \text{cm of Hg},\qquad T_2=-7^{\circ}\text{C}=-7+273=266\ \text{K}.$$

The balloon supports a load of $$185\ \text{kg}$$ under the ground conditions, so $$\text{Load}_1=185\ \text{kg}$$. Now we compute the new load:

$$\text{Load}_2=\text{Load}_1\;\frac{P_2/T_2}{P_1/T_1}.$$

First evaluate each ratio separately:

$$\frac{P_2}{T_2}=\frac{45}{266}=0.169172,$$

$$\frac{P_1}{T_1}=\frac{76}{300}=0.253333.$$

Now form the quotient of these two ratios:

$$\frac{P_2/T_2}{P_1/T_1}=\frac{0.169172}{0.253333}=0.667751.$$

Multiplying this factor by the initial load gives

$$\text{Load}_2=185\times0.667751=123.543\ \text{kg}.$$

Rounding to two decimal places, the new load is $$123.54\ \text{kg}$$.

Hence, the correct answer is Option B.