Question

question 13

the atmospheric pressure at the summit of mt. kanchenjunga - the third highest mountain in the world - is around 455.0 torr.

calculate the volume, in liters, of a balloon at the top of kanchenjunga that has a volume of 11.5 liters at sea level, where the atmospheric pressure is 1.00 atm. assume the temperature differences are negligible.

Explanation:

<pre_analysis>
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"concepts_used": [
"Pressure Unit Conversion"
],
"new_concepts": [
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"Gas Laws"
],
"current_concepts": [
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"Boyles Law",
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<reasoning>

Convert pressure units to a common scale

Using the Pressure Unit Conversion knowledge point
\[

$$\begin{aligned} P_1 &= 1.00\text{ atm} = 760.0\text{ torr} \\ P_2 &= 455.0\text{ torr} \end{aligned}$$

\]

Apply Boyle's Law to find the final volume

\[

$$\begin{aligned} P_1 V_1 &= P_2 V_2 \\ V_2 &= \frac{P_1 V_1}{P_2} \\ V_2 &= \frac{760.0\text{ torr} \times 11.5\text{ L}}{455.0\text{ torr}} \end{aligned}$$

\]

Calculate the final numerical value

\[

$$\begin{aligned} V_2 &= \frac{8740}{455.0}\text{ L} \\ V_2 &\approx 19.2\text{ L} \end{aligned}$$

\]
</reasoning>

<answer>
The atmospheric pressure at the summit of Mt. Kanchenjunga - the third highest mountain in the world - is around 455.0 torr.

Calculate the volume, in liters, of a balloon at the top of Kanchenjunga that has a volume of 11.5 liters at sea level, where the atmospheric pressure is 1.00 atm. Assume the temperature differences are negligible.

<blank>19.2</blank>
</answer>

<post_analysis>
{
"subject": "Natural Science",
"question_type": "Fill-in-the-blank",
"knowledge_point": [
"Natural Science",
"Chemistry",
"Gas Laws"
]
}
</post_analysis>

Answer:

<pre_analysis>
{
"quality": "clear",
"question_count": 1,
"skills_matched": [
"step_cot",
"blank_filling_formatter",
"explain_then_answer",
"quick"
],
"needs_drawing": false,
"concepts_used": [
"Pressure Unit Conversion"
],
"new_concepts": [
"Boyles Law",
"Gas Laws"
],
"current_concepts": [
"Pressure Unit Conversion",
"Boyles Law",
"Gas Laws"
]
}
</pre_analysis>

<reasoning>

Convert pressure units to a common scale

Using the Pressure Unit Conversion knowledge point
\[

$$\begin{aligned} P_1 &= 1.00\text{ atm} = 760.0\text{ torr} \\ P_2 &= 455.0\text{ torr} \end{aligned}$$

\]

Apply Boyle's Law to find the final volume

\[

$$\begin{aligned} P_1 V_1 &= P_2 V_2 \\ V_2 &= \frac{P_1 V_1}{P_2} \\ V_2 &= \frac{760.0\text{ torr} \times 11.5\text{ L}}{455.0\text{ torr}} \end{aligned}$$

\]

Calculate the final numerical value

\[

$$\begin{aligned} V_2 &= \frac{8740}{455.0}\text{ L} \\ V_2 &\approx 19.2\text{ L} \end{aligned}$$

\]
</reasoning>

<answer>
The atmospheric pressure at the summit of Mt. Kanchenjunga - the third highest mountain in the world - is around 455.0 torr.

Calculate the volume, in liters, of a balloon at the top of Kanchenjunga that has a volume of 11.5 liters at sea level, where the atmospheric pressure is 1.00 atm. Assume the temperature differences are negligible.

<blank>19.2</blank>
</answer>

<post_analysis>
{
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"question_type": "Fill-in-the-blank",
"knowledge_point": [
"Natural Science",
"Chemistry",
"Gas Laws"
]
}
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