PV Diagram and Ideal Gas Problem?
PV Diagram and Ideal Gas Problem?
One mole of an ideal gas is confined to a container with a movable piston. The questions below refer to the processes shown on the PV diagram. Process I is a change from state X to state Y at constant pressure. Process II is a change from state Z to state W at a different
constant pressure.
a. Rank the temperatures of states W, X, Y, and Z. If any temperatures are equal, say so explicitly. Explain your reasoning.
b. In the two processes, does the volume of the gas increase, decrease, or remain the
same? Explain your reasoning.
c. Based on your answer to part b, state whether the following quantities are positive,
negative, or zero. Explain your reasoning in each case by referring to a force and a displacement.
i. the work done on the gas during process I.
ii. the work done on the gas during process II.
d. In process I, is the heat transfer to the gas positive, negative, or zero? Explain your reasoning.
***The PV diagram is drawn below
P
l
l Process I
l x o-------------->--------------o y
l
l
l
l
l
l
l Process II
l w o--------------<--------------o z
l__________________________V
Comments
1
Points x and z are at the same temperature
The product Px.Vx = PzVz
2.
In process 1 , volume increases from x to y
In process 2 volume decreases from z to x
.
3
The work done on the gas during process I. is negative since volume increases
the work done on the gas during process II. is positive since volume decreases
4.
volume increases , hear transfer is positive
volume decreases , heat transfer is negative
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PV Diagram and Ideal Gas Problem?
PV Diagram and Ideal Gas Problem?
One mole of an ideal gas is confined to a container with a movable piston. The questions below refer to the processes shown on the PV diagram. Process I is a change from state X to state Y at constant pressure. Process II is a change from state Z to state W at a...
PV = nRT If 'nRT' is held constant, then P & V can vary. If P increass then V decreases and conversely. Thinking of the piston arrangement. depressing the piston decreases to volume but increases the pressure.. The pressure referred to is the internal pressure within the system, that is the pressure within the system. The best everyday analogy is the bicycle pump. When the handle is at its fullest extent the amount of gas is 'n', when the handle is pushed in the amount of gas is still the same, but the pressure within the pump is much higher. Also when pushing the handle in the volume is decreased. A secondary consideration is the temperature.. You will notice that is compressed the gas in the pump, the pump feels hotter. The 'T' value also increases.