We have to consider the ideal gas law, thus the relation between both states will be given by
T1P1V1=T2P2V2⟹V1V2=P2T1P1T2
Now, we need to define the pressures and temperatures to calculate the rate between the new volume out of the well (V2) and the original volume of gas at the bottom (V1):
P1=Pbottom=ρH2OghP1=1000m3kg⋅9.80s2m⋅5m×1kg/(m2⋅s2)1PaP1=49000PaT1=Tbottom=25°C=298.15KP2=Psurface=PatmP2=101528PaT2=Tsurface=27°C=300.15K
Now we substitute and we will find the change in volume:
% of change=100⋅V1V2−V1=100⋅(P2T1P1T2−1) % of change=100⋅((49000 Pa)(298.15 K)(101528 Pa)(300.15 K)−1)∴% of change=100⋅(2.0589−1)=105.89 %
Reference
- Chang, R., & Goldsby, K. A. (2010). Chemistry. Chemistry, 10th ed.; McGraw-Hill Education: New York, NY, USA.
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