# Isothermal expansion

1. Dec 12, 2009

### morsel

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 m3

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
Pi = 400 kPa
Vi = 20 m3
Pf = 100 kPa
Vf = 80 m3

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.

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2. Dec 12, 2009

### rl.bhat

The pressure is kilo pascal.

3. Dec 12, 2009

### morsel

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. Dec 13, 2009

### Andrew Mason

True. But you can blame the person who drafted the question. That is the right answer to this question.

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. Dec 13, 2009