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Isothermal expansion 
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#1
Dec1209, 03:20 PM

P: 32

1. The problem statement, all variables and given/known data
Suppose 161 moles of a monatomic ideal gas undergoes an isothermal expansion as shown in the figure (attached). The horizontal axis is marked in increments on 20 m^{3} What is the temperature at the beginning and at the end of this process? 2. Relevant equations PV = nRT 3. The attempt at a solution P_{i} = 400 kPa V_{i} = 20 m^{3} P_{f} = 100 kPa V_{f} = 80 m^{3} T = (PV)/(nR) = (400*20)/(161*8.31) = 5.98 K My answer is wrong according to the online homework grading system. What am I doing wrong? Any help is appreciated. 


#2
Dec1209, 06:46 PM

HW Helper
P: 4,437

The pressure is kilo pascal.



#3
Dec1209, 07:41 PM

P: 32

OK. But even if I convert kPa to Pa and calculate, that's still not the answer.
T = (400,000 Pa * 20) / (161*8.31) = 5979 K This seems outrageously large. 


#4
Dec1309, 12:39 PM

Sci Advisor
HW Helper
P: 6,654

Isothermal expansion
One mole of a gas at STP occupies 22.4 litres or .0224 m^3. That is at about 100 kPa pressure and a temperature of 273 K. Here, you have 161 times that amount of gas occupying 80/.0224 = 3571 times as much volume. So to reach the same 1 atm pressure the temperature has to be 3571/161 = 22 times the temperature at STP about 6000 degrees. AM 


#5
Dec1309, 07:42 PM

P: 32

Okay. Thanks for your help.



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