A balloon originally has a volume of $1.0 \mathrm {~m} ^ { 3 }$ when the gas in it is at $20 ^ { \circ } \mathrm { C }$ and under a pressure of 1.0 atm. As it rises in the earth's atmosphere, its volume expands. What will be its new volume if its final temperature and pressure are $- 40 ^ { \circ } \mathrm { C }$ and 0.10 atm ?
A) $6.0 \mathrm {~m} ^ { 3 }$
B) $2.0 \mathrm {~m} ^ { 3 }$
C) $4.0 \mathrm {~m} ^ { 3 }$
D) $8.0 \mathrm {~m} ^ { 3 }$

Question

A balloon originally has a volume of $1.0 \mathrm {~m} ^ { 3 }$ when the gas in it is at $20 ^ { \circ } \mathrm { C }$ and under a pressure of  1.0  atm. As it rises in the earth's atmosphere, its volume expands. What will be its new volume if its final temperature and pressure are $- 40 ^ { \circ } \mathrm { C }$ and  0.10 atm ?
A) $6.0 \mathrm {~m} ^ { 3 }$
B) $2.0 \mathrm {~m} ^ { 3 }$
C) $4.0 \mathrm {~m} ^ { 3 }$
D) $8.0 \mathrm {~m} ^ { 3 }$

Quizplus · Accepted Answer

To find the new volume, we can use the combined gas law: $\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}$. Convert temperatures to Kelvin: $T_1 = 20°C + 273.15 = 293.15K$, $T_2 = -40°C + 273.15 = 233.15K$. Solve for $V_2$: $V_2 = \frac{P_1V_1T_2}{P_2T_1} = \frac{1.0 \times 1.0 \times 233.15}{0.10 \times 293.15} \approx 8.0 \mathrm {~m} ^ { 3 }$.

A Balloon Originally Has a Volume Of 1.0 m31.0 \mathrm {~m} ^ { 3 }1.0 m3

A Balloon Originally Has a Volume Of $1.0 \mathrm {~m} ^ { 3 }$