A balloon is filled with 80 liters of gas on a day where the temperature was 34 degrees at sea level which is 101.3 kPa and released. As the balloon rises to a certain altitude, the temperature drops to 0 degrees celsius and the balloon doubles in volume. What is the atmospheric pressure at that altitude?

0.444atm

Explanation:

Using the combined gas law equation;

P1V1/T1 = P2V2/T2

Where;

P1 = initial pressure (

P2 = final pressure (

V1 = initial volume (L)

V2 = final volume (L)

T1 = initial temperature (K)

T2 = final temperature (K)

According to this question,

P1 = 101.3 kPa = 101.3 × 0.00987 = 0.999atm

P2 = ?

V1 = 80L

V2 = 160L (double of V1)

T1 = 34°C = 34 + 273 = 307K

T2 = 0°C = 0 + 273 = 273K

Using P1V1/T1 = P2V2/T2

0.999 × 80/307 = P2 × 160/273

79.92/307 = 160P2/273

Cross multiply

307 × 160P2 = 79.92 × 273

49120P2 = 21818.16

P2 = 21818.16 ÷ 49120

P2 = 0.444

P2 = 0.444atm