InterviewSolution
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InterviewSolution
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A certain gas in expanded from `(1L, 10 atm)` to `(4L, 5 atm)` against a constnt external pressure of `1atm`. If the initial temperature of gas is `300K` and heat capacity for the process is `50J^(@)C^(-1)`, the enthalpy change during the process is: (use: `1L-atm = 100J)`A. `15 kJ`B. `15.7 kJ`C. `14.3 kJ`D. `14.7kJ` |
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Answer» Use: `DeltaH = DeltaU + Delta(PV) = DeltaU + (P_(2)V_(1) - P_(1)V_(1))` `DeltaU` can be calculated as follows: i. Use: `(P_(1)V_(1))/(T_(1)) = (P_(2)V_(2))/(T_(2)) rArr (10 xx 1)/(300) = (5 xx 4)/(T_(2)) rArr T_(2) = 600K` ii. `q = C DeltaT = 50 xx (600 - 300) = 15 kJ` iii. `-w = P_(ex) DeltaV = 1(14-1) = 3 L-atm -= 0.3 kJ` iv. `q = DeltaU +(-w) rArr 15 = DeltaU = 0.3 rArr DeltaU = 14.7 kJ` `rArr DeltaH = 14.7 +(5 xx 4 - 10 xx1) xx 100 xx 10^(-3) = 15.7kJ` |
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