An ideal gas (which is is a hypothetical gas that conforms to the laws governing
ID: 918881 • Letter: A
Question
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 4.90 to 2.45 L . When the external pressure is increased to 2.50 atm, the gas further compresses from 2.45 to 1.96 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 4.90 to 1.96 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?
Explanation / Answer
two -step process : .
W 1-2= pdV = p (V2 - V1)
= 2 (2.45 - 4.90)
= - 4.8 L-atm
W2-3= P (V2 - V1)
= 2.5 (1.96 - 2.45)
= - 1.225 L-atm
total work = W1-2 + W2-3
= -4.8 - 1.225
= - 6.025 L-atm
total work = - 610.33 J
second process :
W = P (V2 - V1)
= 2.50 (1.96 - 4.90)
= - 7.35 L-atm
Work = - 744.5 J
it is isothermal process because final temperature is same for both process.
so dU = Cv dT = 0
q = - w
for two - step process q = 601.33 J
for one - step process q = 744.5 J
by considering this values in one step process the q is more.
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