Asked by Alfred
1.6 mol of Neon gas are heated from 20.0 °C to 85.2 °C at a constant pressure of 1 atm. Calculate
the work done. (R = 0.082057 L.atm/ K. mol)
the work done. (R = 0.082057 L.atm/ K. mol)
Answers
Answered by
drwls
Work done = (Pressure)*(Volume change)
= P(nR(T2-T1)]/P
= n*R(T2-T1)
= 1.6(0.082057 L.atm/ K. mol)*65.2K
= 8.56 L*atm
Those are inconvenient units for work, so multiply by 10^-3 m^2/L*(1.013*10^5 N/m^2/atm
I get 867 N-m (Joules)
= P(nR(T2-T1)]/P
= n*R(T2-T1)
= 1.6(0.082057 L.atm/ K. mol)*65.2K
= 8.56 L*atm
Those are inconvenient units for work, so multiply by 10^-3 m^2/L*(1.013*10^5 N/m^2/atm
I get 867 N-m (Joules)
Answered by
Alfred
so the formula is W = nR∆T
Thank you for the answer above:)
Thank you for the answer above:)
Answered by
Kata
W=-nRT
=-(1.6)(O.O82507)(65)
=-8.53J
=-(1.6)(O.O82507)(65)
=-8.53J
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.