Animated Solution for Physics - Thermodynamics: A balloon carries a total load of $185\text{ kg}$ at normal pressure and temperature of $27^{\circ }\text{C}$. What load will the balloon carry on rising to a height at which the barometric pressure is $45\text{ cm}$ of Hg and the temperature is $-7^{\circ }\text{C}$? [Assuming, the volume constant.]

Imagine a hot air balloon resting on the ground, fully inflated and ready for takeoff. At ground level, the atmospheric pressure is at its normal value—$76\text{ cm}$ of mercury—and the temperature is a comfortable $27^{\circ }\text{C}$ (or $300\text{ K}$). Under these conditions, the balloon can carry a hefty load of $185\text{ kg}$.

But the atmosphere is a dynamic and ever-changing fluid. As the balloon ascends to a high altitude, the environment shifts drastically. The pressure drops to just $45\text{ cm}$ of mercury, and the temperature plummets to a freezing $-7^{\circ }\text{C}$ (or $266\text{ K}$). The question we must answer is: how does this dramatic change in the environment affect the balloon's ability to carry a load, assuming its volume remains perfectly constant?

To solve this mystery, we need to understand the physics of buoyancy. The lifting power of a balloon is governed by Archimedes' principle, which states that the buoyant force is equal to the weight of the displaced fluid. Since the volume of our balloon is constant, the mass of the air it displaces—and therefore the total load it can carry—is directly proportional to the density of the surrounding air.

But how do we calculate the density of the air at different altitudes? We turn to the trusty Ideal Gas Law:

By replacing the number of moles $n$ with the mass $m$ divided by the molar mass $M_w$, we can rewrite the equation to solve for density $\rho$:

This elegant rearrangement reveals a crucial relationship: the density of a gas is directly proportional to its pressure and inversely proportional to its absolute temperature. Mathematically, we write this as:

Since the load $M$ the balloon can carry is directly proportional to the air density $\rho$, we can establish a direct proportionality between the load, pressure, and temperature:

This allows us to set up a powerful ratio comparing the ground state (State 1) to the high-altitude state (State 2):

Simplifying this complex fraction, we get our master equation for the problem:

Now, it is time to carefully substitute our known values into the master equation. We must always remember to use absolute temperature (Kelvin) to avoid mathematical paradoxes!

For the ground state: $p_1=76\text{ cm Hg}$ $T_1=300\text{ K}$

For the high altitude: $p_2=45\text{ cm Hg}$ $T_2=266\text{ K}$

Plugging these into our ratio:

Let's execute the arithmetic. Multiplying the numerators and denominators gives us:

To isolate our unknown final load $M_2$, we simply cross-multiply and rearrange the terms:

Evaluating this final expression yields:

As the balloon rises into the thinner, colder air of the upper atmosphere, the drop in pressure outpaces the drop in temperature, leading to an overall decrease in air density. Consequently, the balloon's lifting capacity drops significantly from $185\text{ kg}$ to just about $123.5\text{ kg}$.

This beautiful interplay of thermodynamics and fluid mechanics perfectly explains why high-altitude balloons must be engineered with such precision. The correct option is (d).