What Is the Highest Temperature Reached in the Ideal Gas Cycle?

[weinbergshaun]

Question (see attached diagram):
PV diagram with 7.5 moles of ideal diatomic gas through cycle a, b and c. What is the highest temperature reached by the gas during the cycle? (multiple choice answers 180, 325, 208 and 100 C, i know answer is 208 C but I'm not getting it!) It is a PV diagram with (triangle), pressure is measured in p(PA x 10^4) and volume is V(m^3) The question is on pg 16 at https://www.scribd.com/doc/287176891:
Point a => p = 3 p(PA x 10^4) and v = 0.2 V(m^3)
Point b => p = 5 p(PA x 10^4) and v = 0.6 V(m^3)
Point c => p = 3 p(PA x 10^4) and v = 0.6 V(m^3)

Attempt at question:
Point a:
p = 3 PA x 10^4 (im keeping this and will use R = 8.31 L.KPA/K.mol I THINK PA x 10^4 = kPA!)
V = 600 Litres (converted cubic metres to litres times by 1000)
n = 7,5 (this is given even though its diatomic, it will remain as 7,5 since its moles and not molar mass?
R = 8.31 L.KPA/K.mol T = ? PV = nRT T = PV/Rn T = 3 x 600/8.31 x 7,5

Using T = nR/PV:
T = 1800/62.32 T = 28.8 Kelvin

If i apply this to all the other points then i get nowhere near the 208 C, so this example above shows that one of my values is wrong (and will be wrong for all other points b and c).

Many thanks for your help!

 

[ehild]

Question (see attached diagram):
PV diagram with 7.5 moles of ideal diatomic gas through cycle a, b and c. What is the highest temperature reached by the gas during the cycle? (multiple choice answers 180, 325, 208 and 100 C, i know answer is 208 C but I'm not getting it!) It is a PV diagram with (triangle), pressure is measured in p(PA x 10^4) and volume is V(m^3)
Attempt at question:
Point a:
p = 3 PA x 10^4 (im keeping this and will use R = 8.31 L.KPA/K.mol I THINK PA x 10^4 = kPA!)

No, "kilo" means 1000. 3 x 10⁴ Pa=30 kPa.
Why do you use liters and kPa-s when the volume is given in m³-s and the pressure in pascals? And R=8.31 m³ Pa/(K mol) .
And the problem asks the temperature in °C.

 

[weinbergshaun]

No, "kilo" means 1000. 3 x 10⁴ Pa=30 kPa.
Why do you use liters and kPa-s when the volume is given in m³-s and the pressure in pascals? And R=8.31 m³ Pa/(K mol) .
And the problem asks the temperature in °C.

Thanks for your assistance. I'm doing the best to figure this out, my books does not give me the laws clearly. So your saying 3 x 10^4 Pa=30 kPa, i was using 3 kPA. I converted cubic meters to liters as i assumed that if you work with pascals you needed to convert the cubic meters to liters and if you work with atmospheric pressure (atm) you work with cubic meters. I also assumed that if you use the kilopascal then you must use R=8.31 and if you just use pascals its the other R = 0,0821 value.

 

[ehild]

Thanks for your assistance. I'm doing the best to figure this out, my books does not give me the laws clearly. So your saying 3 x 10^4 Pa=30 kPa, i was using 3 kPA. I converted cubic meters to liters as i assumed that if you work with pascals you needed to convert the cubic meters to liters and if you work with atmospheric pressure (atm) you work with cubic meters. I also assumed that if you use the kilopascal then you must use R=8.31 and if you just use pascals its the other R = 0,0821 value.

3 x 10^4 Pa=30 kPa, using 3 kPa instead is wrong.
R can be used in different units. See
https://en.wikipedia.org/wiki/Gas_constant.
R=8.31 Pa m³/ (K mol) or R=8.31 kPa L / (K mol) .

 

[weinbergshaun]

Thanks for your assistance. I'm doing the best to figure this out, my books does not give me the laws clearly. So your saying 3 x 10^4 Pa=30 kPa, i was using 3 kPA. I converted cubic meters to liters as i assumed that if you work with pascals you needed to convert the cubic meters to liters and if you work with atmospheric pressure (atm) you work with cubic meters. I also assumed that if you use the kilopascal then you must use R=8.31 and if you just use pascals its the other R = 0,0821 value.

Ok got it, thanks this helps ten millions times i think i can finish the question now~