5 moles of an ideal gas expands isothermally and reversibly from pressure 10 atm to 2 atm at 300k. What is the largers mass which can be lifted through height of 1 m
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Work done in an isothermal reversible process is given by the equation,
-2.303nRTlog(P1/P2) where
n is number of moles
R=8.314J/mol.K
T is temperature
P1,P2 are initial and final pressures.
By substituting, we get
-2.303*5*8.314*300*log(10/2)
-2.303*5*8.314*300*log5 = 20074.91 J
Work done is lifting a mass to a height h is given by m*g*h
-2.303nRTlog(P1/P2) where
n is number of moles
R=8.314J/mol.K
T is temperature
P1,P2 are initial and final pressures.
By substituting, we get
-2.303*5*8.314*300*log(10/2)
-2.303*5*8.314*300*log5 = 20074.91 J
Work done is lifting a mass to a height h is given by m*g*h
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