The figure shows a PV diagram for 2.69 mole of oxygen gas in a sealed

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SUMMARY

The discussion focuses on calculating the temperatures T3 and T4 for 2.69 moles of oxygen gas in a sealed container using the ideal gas law and a PV diagram. The initial temperature (T1) is given as 21°C, and the calculations provided by the user initially yield T2 as 220.5K, which is incorrectly interpreted as 52.5°C due to a sign error. The correct calculation for T4 results in 661.5K, equivalent to 388.5°C, after applying the appropriate formulas for volume and temperature relationships.

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billsauce
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Homework Statement


The figure shows a PV diagram for 2.69 mole of oxygen gas in a sealed container. The temperature of state 1 is 21 C. Pressure is Pi. What is T3 and T4?
http://s74.photobucket.com/user/teysure/media/phyiscs.png.html?sort=3&o=0





my attempt at solving the solution:
T3 - P1V1/T1 = P3V3/T3
(1Pi)(.100l)/294K = (1.5Pi)(.05l)/T2
T2 = 220.5K - 273 = 52.5K

T4 - V3/T3 = V4/T4
(.05l)/(220.5K) = (.150l)/(T4)
T4 = 661.5K = 388.5C
 
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billsauce said:

Homework Statement


The figure shows a PV diagram for 2.69 mole of oxygen gas in a sealed container. The temperature of state 1 is 21 C. Pressure is Pi. What is T3 and T4?
http://s74.photobucket.com/user/teysure/media/phyiscs.png.html?sort=3&o=0my attempt at solving the solution:
T3 - P1V1/T1 = P3V3/T3
(1Pi)(.100l)/294K = (1.5Pi)(.05l)/T2
T2 = 220.5K - 273 = 52.5K

T4 - V3/T3 = V4/T4
(.05l)/(220.5K) = (.150l)/(T4)
T4 = 661.5K = 388.5C
When you calculated T3, you made a sign error. 220.5 - 273 is a negative number, and the result is in degrees °C.