A balloon filled with 0.25 mol of He(g) at 273 K and 1 atm is allowed to rise through the atmosphere — States of Matter Chemistry Question

STEPS:

1. Understand the relationship between pressure, temperature, and volume:
The behavior of helium gas inside the balloon can be modeled using the Ideal Gas Law ($PV = nRT$). Since the amount of helium gas remains constant ($n = 0.25\text{ mol}$) as the balloon rises, we can relate the initial and final states of the gas using the Combined Gas Law:
$$\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}$$
To analyze what happens to the volume, we can rearrange this equation to solve for the final volume ($V_2$) as a function of the initial volume ($V_1$):
$$V_2 = V_1 \times \left(\frac{P_1}{P_2}\right) \times \left(\frac{T_2}{T_1}\right)$$

2. Isolate and calculate the pressure effect:
* The pressure decreases from $1\text{ atm}$ at ground level to $0.1\text{ atm}$ at altitude.
* This is a 10-fold decrease in external pressure ($\frac{P_1}{P_2} = \frac{1\text{ atm}}{0.1\text{ atm}} = 10$).
* According to Boyle's law, gas volume is inversely proportional to pressure. Therefore, this dramatic drop in external pressure acts to increase the volume of the balloon by a factor of 10.

3. Isolate and calculate the temperature effect:
* The absolute temperature of the air decreases from $273\text{ K}$ to $220\text{ K}$ as the balloon rises.
* According to Charles's law, gas volume is directly proportional to absolute temperature. Therefore, this cooling acts to decrease the volume by a factor of:
$$\frac{T_2}{T_1} = \frac{220\text{ K}}{273\text{ K}} \approx 0.81$$
(This corresponds to a $19\%$ contraction in volume).

4. Combine both effects to find the net change:
* Multiply both ratios together to find the net scaling factor for the volume:
$$V_2 = V_1 \times (10) \times (0.81) \approx \mathbf{8.1 \times V_1}$$
* Because $8.1 > 1$, the volume of the balloon must increase.
* The massive expansion caused by the 10-fold drop in pressure completely dominates the relatively small contraction caused by the $19\%$ drop in absolute temperature. This identifies Option A as the correct answer.

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WHY_OTHERS_WRONG:

• Option B is incorrect: The two variables do not counteract each other equally. A 10-fold decrease in pressure (which scales volume by $\times 10$) is orders of magnitude more significant than the temperature decrease (which scales volume by $\times 0.81$).
• Option C is incorrect: This option falsely claims that the temperature decrease has a larger impact on the volume than the pressure decrease. If this were true, the balloon's volume would contract ($V_2 < V_1$), which is mathematically impossible given the data.
• Option D is incorrect: While knowing the initial volume ($V_1$) is required to calculate the exact *numerical* final volume in liters, it is completely unnecessary for determining whether the volume *increases, decreases, or remains the same*. The ratio of the changing variables ($\frac{V_2}{V_1} \approx 8.1$) is constant and proves the balloon expands regardless of its starting size. (Furthermore, the initial volume actually can be calculated using the ideal gas law: $V_1 = \frac{nRT_1}{P_1} = \frac{0.25\text{ mol} \times 0.08206\text{ L atm mol}^{-1}\text{ K}^{-1} \times 273\text{ K}}{1\text{ atm}} \approx 5.6\text{ L}$).