Question

A weather balloon is inflated to a volume of 27.9L at a pressure of 732mmHg and a temperature of 30.1?C. The balloon rises in the atmosphere to an altitude where the pressure is 385mmHg and the temperature is -13.6?C.Assuming the balloon can freely expand, calculate the volume of the balloon at this altitude.

Answer 1

Answer:

45.4 L

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 = 27.9 L

V₂ = ?

P₁ = 732 mmHg

P₂ = 385 mmHg

T₁ = 30.1 ºC

T₂ = -13.6  ºC

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

T(K) = T( °C) + 273.15  

So,  

T₁ = (30.1 + 273.15) K = 303.25 K  

T₂ = (-13.6 + 273.15) K = 259.55 K  

Using above equation as:

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

$\frac{{732}\times {27.9}}{303.25}=\frac{{385}\times {V_2}}{259.55}$

$\frac{385V_2}{259.55}=\frac{20422.8}{303.25}$

$148225V_2=6729708.26018$

Solving for V₂ , we get:

V₂ = 45.4 L