A 2.0-m³ weather balloon is loosely filled with helium at 1 atm (76 cm Hg)and at 27°C.At an elevation of 20,000 ft, the atmospheric pressure is down to 38 cm Hg and the helium has expanded, being under no constraint from the confining bag.If the temperature at this elevation is -48°C, the gas volume is: A)0.75 m³ B)1.3 m³ C)3.0 m³ D)4.0 m³ E)5.3 m³

Question

Quizplus · Accepted Answer

To solve this problem, we can use 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. Given that the initial conditions are 1 atm (76 cm Hg) and 20°C (which is 293K), and the final conditions are 38 cm Hg and -48°C (-48°C = 225K), we can solve for $V_2$. The initial volume $V_1$ is 2.0 m³. Converting pressures to the same units and temperatures to Kelvin, we can solve for $V_2$, which results in approximately 3.0 m³.

A 20-M3 Weather Balloon Is Loosely Filled with Helium at 1

A 20-M³ Weather Balloon Is Loosely Filled with Helium at 1