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Plz answer this.. An open flask contains air at 27°C. Calculate the fraction of air that would be expelled at 476°C

Plz answer this..
An open flask contains air at 27°C. Calculate the fraction of air that would be expelled at 476°C

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Grade:11

2 Answers

Arun
25750 Points
4 years ago
Initial Pressure (P1) = 1 atm
Initail Temperature (T1) = 27*C = 300 K
Final Pressure (P2) = P1 = 1 atm
​Final Temperature (T2) = 127*C = 400 K

Initial Volume (V1) = V
Final Volume (V2) = V'

Using Combined Gas Equation, we get:
P1V1T1 = P2V2/T2
​Substituting the values, we get: V' = 4V/3 

But, the flask can hold only V volume of air.This makes us realise that some air has been expelled out of the flask.
=> Volume of air expelled = 4V/3 - V = V/3

From this, we can get the fraction of air expelled out = Amount of air expelled/Final volume
=>  Fraction of air expelled out = V/3 / 4V/3 = 1/4
Vikas TU
14149 Points
4 years ago
Dear student 
Here is the assumptions and detailed explanation 
Assumptions :
Volume of flask is constant
Air at 27 °C & 1 atm is behave like Ideal Gas.
Expansion of flask due to heat in nil
So, if gas to be expel 1/4th of Volume (it is final Volume, not initial) then it has to expand in sich way that it's final volume would be Volume of flask + 1/4 th of Gas (final volume).
Assume, Volume of flask = V
Initial Conditions,
T = 27 °C = 300 K
Volume = V
Final Conditions,
Final Temp = T”
Final Volume (V') = V + 1/4 of V'
V' = V + V'/4
V' - V'/4 = V
3V’/4 = V
V' = 4V/3
From ideal gas equation, we know that volume and temperature are directly proportional.
V/T = V'/T”
T” = T × V'/V
T” = T × 4/3V/V
T” = T × (4/3)
T” = (4/3)×300
T” = 400K = 127 °C
So, at 127 °C, gas will expel it's 1/4th volume out of flask.
Final volume will not be equal to 5/4 time of initial volume, because in question it mentioned that, it expel 1/4 of it's gas volume, not flask volume. So it will expel 1/4 of it's final volume.

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