Solving Isothermal Expansion Problem - Temperature Calculation

[morsel]

Homework Statement

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?

Homework Equations

PV = nRT

The Attempt at a Solution

P_i = 400 kPa
V_i = 20 m³
P_f = 100 kPa
V_f = 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.

 

[rl.bhat]

The pressure is kilo pascal.

 

[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.

 

[Andrew Mason]

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.

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

 

[morsel]

Okay. Thanks for your help.