Question

A gas-filled weather balloon with a volume of 65.0 L is released at sea level where conditions are745 torr and 25 ºC. The balloon can expand to a maximum volume of 835 L. When the balloon rises to an altitude at which the temperature is –5 ºC and the pressure is 0.066 atm, will it have exceeded its maximum volume?

Answer 1

Answer:

It will reach to its maximum volume.

Explanation:

Using Ideal gas equation for same mole of gas as :-

$\frac {{P_1}\times {V_1}}{T_1}=\frac {{P_2}\times {V_2}}{T_2}$

Given ,  

V₁ = 65.0 mL

V₂ = ?

P₁ = 745 torr

The conversion of P(torr) to P(atm) is shown below:

$P(torr)=\frac {1}{760}\times P(atm)$

So,  

Pressure = 745 / 760 atm = 0.9803 atm

P₁ = 0.9803 atm

P₂ = 0.066 atm

T₁ = 25 ºC

T₂ = -5 ºC

The conversion of T( °C) to T(K) is shown below:

T(K) = T( °C) + 273.15  

So,  

T₁ = (25 + 273.15) K = 298.15 K  

T₂ = (-5 + 273.15) K = 268.15 K  

Using above equation as:

$\frac {{P_1}\times {V_1}}{T_1}=\frac {{P_2}\times {V_2}}{T_2}$

$\frac{{0.9803}\times {65.0}}{298.15}=\frac{{0.066}\times {V_2}}{268.15}$

Solving for V₂ as:-

$V_2=\frac{0.9803\times \:65.0\times 268.15}{298.15\times 0.066}$

$V_2=\frac{17086.38392}{19.6779}$

V₂ = 868 L

Given that:- V max = 835 L

Thus, it will reach to its maximum volume.