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Time management is very much important in IIT JAM. The eduncle test series for IIT JAM Mathematical Statistics helped me a lot in this portion. I am very thankful to the test series I bought from eduncle.
Nilanjan Bhowmick AIR 3, CSIR NET (Earth Science)
Shweta Thakur posted an Question
June 01, 2020 • 20:11 pm
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30 points
30 points
- IIT JAM
- Chemistry (CY)
Plz explain this ... u calories of heat are absorbed by the system. the molar volumes of ice and ualer
U Calories of heat are absorbed by the system. The molar volumes of ice and Ualer are 0-0196 and 0-0180 litre respectively. Calculate AH and AE. Solution: Since AH = p bx. 5. When 1 mole of ice melts at 0°C and at constant pressure of 1 atm, heat absorbed by the system at constant pressure = 1440 calories. In the equ uation AH = AU + pAV pAV=76x 13.6 x 981 (18-19.6) ergs 76x 13-6 x 981 x 1-6 4-18 x 10 pAV is very small compared to AU, pAV can be neglected. calories p-1 atm 76 x 13-6 x 981 dynes, hus, AH= AU 1440 calories. =-0-039 calories. V2 = 18 cc, Vi = 19.6 cc) Snce
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Lingareddy 1
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pressure 1atm= 76 cm of hg(mercury ) Mercury (Hg)is 13.6 times denser than water acceleration due to gravity =981cm/s2
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Anand pratap singh
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Follow the image
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Dinesh khalmaniya 1
conversion of 1 atm
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Shweta thakur
?
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Shweta thakur
and sir the question which I have sent u there in the solution they have convert work into calories by multiplying 13.6*76*981*volume in cc why
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Shweta thakur
why there is negligible work done here
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work done is given by w = -pdv right? , here given that delta H = 1440 and we got delta U = 1439.97 so here work done will be delta U - delta H = 1440 - 1439.97 = .03 cal. it is very small quantity so we can neglected work done
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so delta H = delta U
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ok sir
Shweta thakur
thanku teachers