A weather balloon containing $2.0 \mathrm {~m} ^ { 3 }$ of hydrogen gas rises from a location at which the temperature is $22 ^ { \circ } \mathrm { C }$ and the pressure is 101 kPa to a location where the temperature is $- 39 ^ { \circ } \mathrm { C }$ and the pressure is 20 kPa. If the balloon is free to expand so that the pressure of the gas inside is equal to the ambient pressure, what is the new volume of the balloon?
A) $8.0 \mathrm {~m} ^ { 3 }$
B) $4.0 \mathrm {~m} ^ { 3 }$
C) $12 m ^ { 3 }$
D) $10 m ^ { 3 }$
E) $6.0 \mathrm {~m} ^ { 3 }$

Question

A weather balloon containing $2.0 \mathrm {~m} ^ { 3 }$ of hydrogen gas rises from a location at which the temperature is $22 ^ { \circ } \mathrm { C }$ and the pressure is  101 kPa to a location where the temperature is $- 39 ^ { \circ } \mathrm { C }$ and the pressure is  20 kPa. If the balloon is free to expand so that the pressure of the gas inside is equal to the ambient pressure, what is the new volume of the balloon?
A) $8.0 \mathrm {~m} ^ { 3 }$
B) $4.0 \mathrm {~m} ^ { 3 }$
C) $12 m ^ { 3 }$
D) $10 m ^ { 3 }$
E) $6.0 \mathrm {~m} ^ { 3 }$

Quizplus · Accepted Answer

The problem can be solved using the combined gas law, which is $P_1V_1/T_1 = P_2V_2/T_2$, where $P$ is pressure, $V$ is volume, and $T$ is temperature in Kelvin. First, convert temperatures to Kelvin: $22^\circ C = 295 K$ and $-39^\circ C = 234 K$. Then, plug in the given values: $101 \times 2.0 / 295 = 20 \times V_2 / 234$. Solving for $V_2$ gives $V_2 = 8.0 \mathrm{m}^3$.

A Weather Balloon Containing 2.0 m32.0 \mathrm {~m} ^ { 3 }2.0 m3

A Weather Balloon Containing $2.0 \mathrm {~m} ^ { 3 }$