A machine employs the isothermal expansion of 1 mol of an ideal gas from 4.50 L to 15.0 L. At 25°C, the machine performs 1.5 kJ of work.?

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[匿名]

2017-03-23 4:43 pm

What percent of the maximum possible work is the machine
producing?

更新1:

A machine employs the isothermal expansion of 1 mol of an ideal gas from 4.50 L to 15.0 L. At 25°C, the machine performs 1.5 kJ of work.

回答 (1)

NCS

2017-03-24 4:02 am

Work = n*R*T*ln(Vf/Vi)
from the citation below. So
Work = 1mol * 8.314J/mol·K * 298K * ln(15.0/4.50)
Work = 2983 J
and so the machine is doing
1500J/2983J * 100% = 50.3% ◄
of the maximum possible.

參考: http://www.ux1.eiu.edu/~cfadd/1150/14Thermo/work.html

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