Thermodynamic problem, mastering physics

AI Thread Summary
The discussion centers on calculating the work required to compress helium gas in a container under two conditions: constant pressure and constant temperature. For constant pressure, the initial calculation yielded a work value of 158.34, but there was confusion regarding unit consistency and the proper evaluation of the integral. The correct approach involves integrating the pressure over the change in volume, which was initially overlooked. After reevaluating the calculations and addressing unit errors, the correct answers were obtained. The importance of careful unit management and proper integration techniques in thermodynamic problems is emphasized.
Shayna
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Homework Statement
A 3000 cm3 container holds 0.10 mol of helium gas at 300C.
How much work must be done to compress the gas to 1000 cm3 at constant pressure?
How much work must be done to compress the gas to 1000 cm3 at constant temperature?


The attempt at a solution
W = integration p dv
For constant pressure,
because pressure is constant, therefore it matters not what integral it is on
W= p = nRT/v = 0.1*8.314*573.15/3 = 158.34


For constant temperature,
W= integration p dv over integral v=1L and v=3L
W= 317.6

apparently Mastering physics thinks this is neither here nor there, well, I am very :confused:
 
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Hi Shayna,

Shayna said:
Homework Statement
A 3000 cm3 container holds 0.10 mol of helium gas at 300C.
How much work must be done to compress the gas to 1000 cm3 at constant pressure?
How much work must be done to compress the gas to 1000 cm3 at constant temperature?


The attempt at a solution
W = integration p dv
For constant pressure,
because pressure is constant, therefore it matters not what integral it is on
W= p = nRT/v = 0.1*8.314*573.15/3 = 158.34

Looking at the first one, this does not look right to me. The work is not equal to the pressure (for one thing they don't have the same units).

You need to evaluate the integral:

<br /> W = - p \int\limits_{V_i}^{V_f} dV<br />

In your work you have determined p, but the volume integral is not equal to 1, so there is more to the answer than just p. What do you get?

(Also, be careful with your units.)
 
Okay, I start from scratch again and got the right answer
I've made multiple mistakes due to carelessness in term of unit.

Thanks so much
 
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