Calculating Work, Volume, and Energy Change for Isothermal Compression

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The discussion focuses on calculating the final volume, work done, and energy change for a 2.00-mol sample of helium gas undergoing isothermal compression. The initial calculations yielded incorrect results due to unit conversion errors, particularly with pressure and the gas constant R. The correct final volume is determined to be 0.0410 m³, with work done on the gas calculated at 5.48 kJ, and the energy change also at -5.48 kJ. Participants emphasize the importance of consistent unit usage, particularly converting atm to Pa for accurate results. The thread concludes with a clarification on the correct mole conversion and the significance of checking calculations for accuracy.
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Homework Statement


A 2.00-mol sample of helium gas initially at 300K and 0.400 atm is compressed isothermally to 1.20atm. Assume the helium behaves as an ideal gas, find (a) the final volume of the gas, (b) the work done on the gas, and (c) the energy temperature?

Homework Equations


PV = nRT

W = nRTln(\frac{Pi}{Pf})

\delta{U} = Q - W

The Attempt at a Solution


This is what i have done can anyone tell me what is wrong, because it is not the same as the book's answer.
a)
Vf = \frac{2 * 8.314*10^3 * 300}{1.20}

Vf = 4157000m^3b)
W = 2 * 8.314 * 10^3 * 300 * ln(\frac{1.20}{0.400})
W = 5480317.541

c)
\delta{U} = -5480317.541 meaning this is compression

The correct answers for the following are:

a) 0.0410 m^3
b) 5.48 kJ
c) -5.48 kJ

P.S
 
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It's the classic error: not checking your units.
 
could you tell me specifically which values have the wrong units?
 
Check your units for "R"
 
how do you convert mol into kmol??
 
ok would 2mol be 0.200 kmol because 1 mol is 1000mols?
 
yes because 1000 moles= 1 kmol
 
after i changed the units i still didn't get the correct answer:

This is what i changed:

Vf = \frac{.200 * 8.314*10^3 * 300}{1.20}
Vf = 415700m^3
 
Paymemoney said:
after i changed the units i still didn't get the correct answer:

This is what i changed:

Vf = \frac{.200 * 8.314*10^3 * 300}{1.20}
Vf = 415700m^3
This is an enormous volume and obviously wrong!

What are your units for pressure?

What is the initial volume? Since T is unchanged, what is the final volume if the pressure is 3 x greater?

AM
 
  • #10
It's .002 Kmole, change 1.2 atm to units of pressure in N/m^2 and 1 J = 1 N*m
 
  • #11
I am not sure why you are using Kmol. It is much easier to use moles and R = 8.314 (J/mol - K) and pressure as J/m^3 or N/m^2

AM
 
  • #12
My reference uses R= 8.314 kJ/kmole-K same as 8.314 J/mole-K pick your poison but keep the units consistent
 
  • #13
Paymemoney said:
Vf = \frac{.200 * 8.314*10^3 * 300}{1.20 * 1.013*10^5}
Vf = 0.0410m^3

i changed the units of atm to Pa and i got the correct answer.
 
  • #14
Paymemoney said:
i changed the units of atm to Pa and i got the correct answer.
I trust that you used .002Kmol not .2Kmol. as you have shown.

AM
 
  • #15
oh yeh oops typo.
 
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