Answer:
45.4 L
Explanation:
Using Ideal gas equation for same mole of gas as
[tex] \frac {{P_1}\times {V_1}}{T_1}=\frac {{P_2}\times {V_2}}{T_2}[/tex]
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:
[tex] \frac {{P_1}\times {V_1}}{T_1}=\frac {{P_2}\times {V_2}}{T_2}[/tex]
[tex]\frac{{732}\times {27.9}}{303.25}=\frac{{385}\times {V_2}}{259.55}[/tex]
[tex]\frac{385V_2}{259.55}=\frac{20422.8}{303.25}[/tex]
[tex]148225V_2=6729708.26018[/tex]
Solving for V₂ , we get:
V₂ = 45.4 L
The volume of the balloon at this altitude where the pressure is 385mmHg and the temperature is -13.6 is 45.4L
The relationship between the pressure, volume and temperature of a gas is given by:
[tex]\frac{P_1V_1}{T_1}= \frac{P_2V_2}{T_2}[/tex]
Given that P₁ = 732 mmHg, V₁ = 27.9, T₁ = 30.1⁰C = 303.25K, P₂ = 385 mmHg, V₂ , T₂ = -13.6⁰C = 259.55K
Hence, using the equation:
[tex]\frac{P_1V_1}{T_1}= \frac{P_2V_2}{T_2}\\\\\frac{732*27.9}{303.25} =\frac{385*V_2}{259.55} \\\\V_2=45.4L[/tex]
Hence the volume of the balloon at this altitude where the pressure is 385mmHg and the temperature is -13.6 is 45.4L
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