Compressing ideal gas isothermally, calc work

AI Thread Summary
The discussion focuses on calculating the work done on an ideal gas during isothermal and reversible compression from 1 to 10 atmospheres at 300K. The relevant equations include the work done equation dw = -PdV and the ideal gas law PV = nRT. Participants suggest using the Clapeyron equation to determine initial and final volumes, allowing for the integration of pressure over the volume change. It is noted that pressure is not constant during the process, but can be derived from the ideal gas law for integration. The final calculation involves integrating to find the work done, which results in a natural logarithm expression.
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Homework Statement



10 moles of an ideal gas are compressed isothermally and reversibly from 1 to 10 atmospheres at 300K. Determine the work done ON the gas.

Homework Equations



dw=-PdV
PV=nRT

The Attempt at a Solution



dT=0
T=300K
dP=10atm

calc dV from ideal gas law = 2.27e10 m^3

so now we have dV for work eqn, but P is not constant?

Thanks.
 
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As much as I don't like thermodynamics, it doesn't seem very complicated.
Start with claperyon equation nr. 1:
p1V1=nRT
which allows you to calculate the first volume.
Equation nr.2:
p2V2=nRT
allows you to calculate 2nd volume. Hence you've got your integrating limits.
Now, I wish I knew how to use TeX in here :/.
Anyway,
<br /> W=\int\limits_{v_1}^{v_2} pdV
As you mentioned, p is not constant, but you can, again, calculate it easily from clapeyron's equation:
pV=nRT. And then substitute p under integral with what you've got, integrate. Should get natural logarithm.
 
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Thank you, irycio!
 
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