Respuesta :
The number of moles will remain the same in the two cases
so we can equate using ideal gas equation which is PV =nRT
Where
P = Pressure
V = volume
n = moles
R = gas constant
T = temperature
for the two conditions
[tex]\frac{P_{1}V_{1}}{T_{1}}=\frac{P_{2}V_{2}}{T_{2}}[/tex]
Putting values
[tex]\frac{P_{1}V_{1}}{T_{1}} =\frac{150X20}{300}=\frac{P_{2}V_{2}}{T_{2}}=\frac{1XV_{2}}{310}[/tex]
Thus
V2 = 3100 L
Explanation:
The given data is as follows.
[tex]V_{1}[/tex] = 20 liter, [tex]V_{2}[/tex] = ?
[tex]P_{1}[/tex] = 150 atm, [tex]P_{2}[/tex] = 1.00 atm
[tex]T_{1}[/tex] = 27 + 273 = 300 K, [tex]T_{2}[/tex] = 37 + 273 = 310 K
Therefore, calculate the value of [tex]V_{2}[/tex] as follows
[tex]\frac{P_{1}V_{1}}{T_{1}}[/tex] = [tex]\frac{P_{2}V_{2}}{T_{2}}[/tex]
[tex]\frac{150 atm \times 20 liter}{300 K}[/tex] = [tex]\frac{1.00 atm \times V_{2}}{310 K}}[/tex]
[tex]V_{2} = \frac{10 atm L K^{-1} \times 310 K}{1.00 atm}[/tex]
= 3100 liter
Therefore, we can conclude that the volume of the balloon is 3100 liter.
