An ideal gas (which is is a hypothetical gas that conforms to the laws governing gas behavior) confined to a container with a massless piston at the top. (Figure 2) A massless wire is attached to the piston. When an external pressure of 2.00 atm is applied to the wire, the gas compresses from 6.60 to 3.30 L . When the external pressure is increased to 2.50 atm, the gas further compresses from 3.30 to 2.64 L . In a separate experiment with the same initial conditions, a pressure of 2.50 atm was applied to the ideal gas, decreasing its volume from 6.60 to 2.64 L in one step. If the final temperature was the same for both processes, what is the difference between q for the two-step process and q for the one-step process in joules?
q in two step process:
First we have to calculate workdone.
w = -PdV = -P(V2-V1)
w = w1+w2
= [-2.0 atm (6.6 L-3.3 L)] + [-2.5 atm (3.3 L - 2.64 L) ]
= -8.25 L.atm
= -8.25 x 101.325 J [ 1 L.atm = 101.325 J]
= -836 J
w = -836 J
From 1st law of thermodynamics,
dU = q + w
Since Temperature is constant, dU = 0
q+ w =0
q = -w = - ( -836 J)
q = + 836 J
Therefore,
q in two step process = + 836 J
q in one step process:
First we have to calculate workdone.
w = -PdV = -P(V2-V1)
= -2.5 atm (6.6 L-2.64 L)]
= - 9.9 L.atm
= -9.9 x 101.325 J [ 1 L.atm = 101.325 J]
= -1003 J
w = -1003 J
From 1st law of thermodynamics,
dU = q + w
Since Temperature is constant, dU = 0
q+ w =0
q = -w = - ( -1003 J)
q = + 1003 J
Therefore,
q in one step process = + 1003 J
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Difference between q for the two-step process and q for the one-step process
= 836 J - 1003 J
= -167 J
An ideal gas (which is is a hypothetical gas that conforms to the laws governing gas...