Question

A gas expands from I to F in the figure below. The energy added to the gas by heat is 444 J when the gas goes from I to F along the diagonal path.Three paths are plotted on a PV diagram, which has a horizontal axis labeled V(liters), and a vertical axis labeled P (atm).The green path starts at point I (2,4), extends vertically down to point B(2,1), then extends horizontally to point F (4,1).The blue path starts at point I (2,4), and extends down and to the right to end at point F (4,1).The orange path starts at point I(2,4), extends horizontally to the right to point A (4,4), then extends vertically down to end at point F(4,1).(a) What is the change in internal energy of the gas?J(b) How much energy must be added to the gas by heat for the indirect path IAF to give the same change in internal energy?J

Answer 1

Option a) - The change in internal energy = -88.5 J

Option b) - Energy that must be added to the gas by heat for the indirect path IAF to give the same change in internal energy = 722 J

According to the data given in the question,

The energy added to gas by heat = 444 J.

The path of the gas = from I to F diagonally.

a) The change in internal energy of the gas,

Work done by the gas as it moves along a direct path IF,

W = - ( - the area that is under the curve )

W = - [ ( 1.00 atm ) (4L - 2L) + $\frac{1}{2}$(4 - 1)(4 - 2) ]

W = - 5 atmL

The change in internal energy is given as = Δu

Δu = Q + W

Δu = 444 J - 5 atm.L

Δu = - $[ \frac{1.019 * 10^{3} }{1 atm} ] [\frac{10^{-4} * m^{4} }{1L} ]$

Δu = -88.5 J

b) Energy to be added = Q

W = -(4) (4 -2) $[ \frac{1.019 * 10^{3} }{1 atm} ] [\frac{10^{-4} * m^{3} }{1L} ]$

W = -810 J

By the usage of Thermodynamics's first law-

Q = Δu - W

Q = -88.5 J - ( - 810 J )

Q = 722 J

Therefore,

Option a) - The change in internal energy = -88.5 J

Option b) - Energy that must be added to the gas by heat for the indirect path IAF to give the same change in internal energy = 722 J

To learn more about the law of thermodynamics,

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