Answer to Question #103116 in General Chemistry for aeriel

Question #103116
How will u compare the amount of dissolved oxygen in a mountain lake at 5,000 ft and 18°C with that in a lake near sea level at the same temperature? Explain.
1
Expert's answer
2020-02-18T06:37:09-0500

Pressure at the sea level: 1 atm.

Pressure at 5000 ft: 0.836 atm. Why? Write hypsometric equation:


p2=p1 exp[g(Z1Z2)RdT]= =1 exp[9.81(05000 ft0.3048 m/ft)287.04(18+273.15)]=0.836 atm.p_2=p_1\text{ exp}\bigg[\frac{g(Z_1-Z_2)}{R_dT}\bigg]=\\ \space\\ =1\text{ exp}\bigg[\frac{9.81(0-5000\text{ ft}\cdot0.3048\text{ m/ft})}{287.04\cdot(18+273.15)}\bigg]=0.836\text{ atm}.

Now refer to Tromans (1998) to calculate the molal concentration of oxygen dissolved in water:


caq=PO2exp[0.046T2+203.35T ln(T/298)(29.378+0.092T)(T298)205918.3144T],c_\text{aq}=P_{\text{O}_2}\text{exp}\bigg[\frac{0.046T^2+203.35T\text{ ln}(T/298)-(29.378+0.092T)(T-298)-20591}{8.3144T}\bigg],

where temperature is in kelvins, pressure is in atm:

caq(0 ft)=1.447103 mol/kg,caq(5000 ft)=1.210103 mol/kg.c_\text{aq}(0\text{ ft})=1.447\cdot10^{-3}\text{ mol/kg},\\ c_\text{aq}(5000\text{ ft})=1.210\cdot10^{-3}\text{ mol/kg}.

References:

Tromans, D. (1998). Temperature and pressure dependent solubility of oxygen in water: a thermodynamic analysis. Hydrometallurgy48(3), 327-342. doi: 10.1016/s0304-386x(98)00007-3


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