Question #60815

the south pole of mars is covered with dry ice, which partly sublimes during the summer.The CO2 vapor recondenses in the winter when the temperature drops to 150k. Given that the heat sublimation of CO2 is 25.9kJ/mol,calculate the atmospheric pressure on the surface of mars.
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Expert's answer

2016-07-14T11:44:02-0400

Answer on Question #60815, Chemistry / General Chemistry

Conditions: the south pole of mars is covered with dry ice, which partly sublimes during the summer. The CO2 vapor recondenses in the winter when the temperature drops to 150K. Given that the heat sublimation of CO2 is 25.9 kJ/mol, calculate the atmospheric pressure on the surface of mars.

The same type of equation used for determination of heats of vaporizations applies to heats of sublimation. The South Pole of Mars is covered with dry ice, which partially sublimes during the summer. The CO2 vapor recondenses in the winter when the temperature drops to 150 K. Given that the heat of sublimation of CO2 is 25.9 kJ/mole, calculate the atmospheric pressure on the surface of Mars if the normal sublimation temperature of dry ice is -75C. Show all work clearly for full credit.


Hs(CO2)=25.9kJ/mol\mathrm{H}_s(\mathrm{CO}_2) = 25.9\,\mathrm{kJ/mol}T1subj(CO2)=75C=273.1578.5=198.15K\mathrm{T}_{1\text{subj}}(\mathrm{CO}_2) = -75\,\mathrm{C} = 273.15 - 78.5 = 198.15\,\mathrm{K}T2=150K\mathrm{T}_2 = 150\,\mathrm{K}


P-?


ΔT=198.15150=48.15(K)\Delta T = 198.15 - 150 = 48.15\,\mathrm{(K)}


Here V1=V2V_1 = V_2

Clausius - Clapeyron in differential form is written:


dP/dT=PΔHf.p./RT2\mathrm{d}P / \mathrm{d}T = P \cdot \Delta H f.p. / RT_2


or


ln(P2/P1)=ΔHϕ,n(T2T1)/RT2T1\ln(P_2 / P_1) = \Delta H_{\phi, n} \cdot (T_2 - T_1) / RT_2 T_1ln(P2/P1)=25.9kJ/mol(48.15K)/8.31(kJ/molK)198.15K150K=0.005\ln(P_2 / P_1) = 25.9\,\mathrm{kJ/mol} \cdot (48.15\,\mathrm{K}) / 8.31\,\mathrm{(kJ/mol\cdot K)} \cdot 198.15\,\mathrm{K} \cdot 150\,\mathrm{K} = 0.005P2/P1=e0.005=1.049(times)P_2 / P_1 = e^{0.005} = 1.049\,(\text{times})CO2CO2,H=25.9kJ/mol\mathrm{CO}_2 \rightarrow \mathrm{CO}_2, \quad H = 25.9\,\mathrm{kJ/mol}P(Mars)=0,71,155kPaP(\text{Mars}) = 0,7 - 1,155\,\mathrm{kPa}Here of 1.1551.049(t)=1.21kPa\text{Here of } 1.155 \cdot 1.049(t) = 1.21\,\mathrm{kPa}


Answer: the atmospheric pressure on the surface of mars 1.21 kPa

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