Question

A balloon filled with helium gas has a volume of 500 mL at a pressure of 1 atm. The balloon is released and reaches an altitude of 6.5 km, where the pressure is 0.5 atm. If the temperature has remained the same, what volume does the gas occupy at this height?

Answer 1

We are using the General gas equation P x V/K = P x V/K
1. P = 1atm V=500ml so PxV= 500 at 6.5km P = 0.5atm V = ? so P xV = 0.5 x V
(We don't have to worry about temperature!) 500 = 0.5 x V so V = 1000ml

2. NO CHANGE in pressure here so we have V/K V=2.75 K = 20 + 273=293 so V/K= 2.75/293
Next set V = 2.46 K = ? so V/K = 2.46/K then 2.75/293= 2.46/K so K=(293/2.75)x2.46
=262 K
Convert back to Celsius 262 - 273 = -11 C
It's raining so I have to rescue the laundry!
Laundry rescued!

3.Now we have to use all three variables. I am using 273K and 100kPa for STP.
P = 100 V = 700 K = 273 These are altered P - unknown, V = 200 K = 273+30=303
!00 x 700/273 = 256.4 this is equal to P x 200/303 = P x 0.66
so P = 256.4/0.66 = 388.48kPa

Answer 2

Answer:Volume of the gas occupy at this height will be 1 L.

Explanation:

Volume of the balloon when pressure of helium gas was 1 atm :

= 500 mL = 0.5 L =$V_1$ (1000 mL= 1 L)

Volume of the balloon when the pressure of helium gas is 0.5 atm :

V=$V_2$

According to Boyle's Law:

$Pressure\propto \frac{1}{Volume}$ (At constant temperature)

$P_1V_1=P_2V_2$

$P_1=1 atm, P_2=0.5 atm$

$1 atm\times 0.5 L=0.5 atm\times V$

$V=\frac{1 atm\tiomes 0.5 L}{0.5 atm}= 1 L$

Volume of the gas occupy at this height will be 1 L.