What is the approximate concentration of carbon dioxide in the atmosphere? Select one: 1. 380 parts per million (0.038%) 2. 7300
1 answer:
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Use the Clausius-Clapeyron equation...
<span>Let T1 be the normal boiling point, which will occur at standard pressure (P1), which is 101.3 kPa (aka 760 torr or 1.00 atm). You know the vapour pressure (P2) at a different temperature (T2). And you are given the enthalpy of vaporization. Therefore, we can use the Clausius-Clapeyron equation.
</span>![ln(P_1/P_2) = \frac{-\delta H_{vap}}{R} \times [\frac{1}{T_1} - \frac{1}{T_2}]](https://tex.z-dn.net/?f=ln%28P_1%2FP_2%29%20%3D%20%5Cfrac%7B-%5Cdelta%20H_%7Bvap%7D%7D%7BR%7D%20%20%5Ctimes%20%5B%5Cfrac%7B1%7D%7BT_1%7D%20-%20%5Cfrac%7B1%7D%7BT_2%7D%5D)
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</span><span>ln(101.3 kPa / 52.7 kPa) = (-29.82 kJ/mol / 8.314x10^{-3} kJ/molK) (1/T - 1/329 K)
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------ some algebra goes here -----
<span>T = 349.99K ...... or ...... 76.8C </span>
<span>Let T1 be the normal boiling point, which will occur at standard pressure (P1), which is 101.3 kPa (aka 760 torr or 1.00 atm). You know the vapour pressure (P2) at a different temperature (T2). And you are given the enthalpy of vaporization. Therefore, we can use the Clausius-Clapeyron equation.
</span>
</span><span>ln(101.3 kPa / 52.7 kPa) = (-29.82 kJ/mol / 8.314x10^{-3} kJ/molK) (1/T - 1/329 K)
</span>
------ some algebra goes here -----
<span>T = 349.99K ...... or ...... 76.8C </span>