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