A weather balloon is loosely inflated with helium at a pressure of 1.00 atm (=76.0 cm Hg) and a temperature of 22.0ºC. The gas volume is 3.47 m³ . At an elevation of 6.50 km, the atmospheric pressure is down to 36.0 cm Hg, and the helium has expanded, being under no restraint from the confining bag. At this elevation the gas temperature is _48.0ºC. What is the gas volume now?

To determine the new volume of the helium gas in the weather balloon at an elevation of 6.50 km, we can apply the combined gas law, which relates pressure, volume, and temperature of a gas. The combined gas law is expressed as:

P1 × V1 / T1 = P2 × V2 / T2

Where:

• P1 = initial pressure
• V1 = initial volume
• T1 = initial temperature (in Kelvin)
• P2 = final pressure
• V2 = final volume
• T2 = final temperature (in Kelvin)

First, we need to convert all temperatures from Celsius to Kelvin and pressures from cm Hg to atm for consistency.

Step 1: Convert Temperatures

The conversion from Celsius to Kelvin is done by adding 273.15:

• T1 = 22.0ºC + 273.15 = 295.15 K
• T2 = -48.0ºC + 273.15 = 225.15 K

Step 2: Convert Pressures

Next, we convert the pressures from cm Hg to atm. The conversion factor is 1 atm = 76.0 cm Hg:

• P1 = 1.00 atm
• P2 = 36.0 cm Hg × (1 atm / 76.0 cm Hg) = 0.4737 atm

Step 3: Substitute Values into the Combined Gas Law

Now we can plug in the values into the combined gas law:

1.00 atm × 3.47 m³ / 295.15 K = 0.4737 atm × V2 / 225.15 K

Step 4: Solve for V2

Rearranging the equation to solve for V2 gives us:

V2 = (1.00 atm × 3.47 m³ × 225.15 K) / (0.4737 atm × 295.15 K)

Step 5: Calculate V2

Now, we can perform the calculations:

• Numerator: 1.00 × 3.47 × 225.15 = 780.56
• Denominator: 0.4737 × 295.15 = 139.59

Thus, V2 = 780.56 / 139.59 ≈ 5.59 m³

Final Result

At an elevation of 6.50 km, the volume of the helium gas in the weather balloon expands to approximately 5.59 m³.