delta U [ENDORSED]
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Re: delta U
Yes, like Mitch said above, delta U has two components, as seen in the equation deltaU = q + w (the first law of thermodynamics). Even if change in temperature is 0, and q is 0, work can still be done and affect internal energy.
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Re: delta U [ENDORSED]
If there is no change in temperature in an ideal gas then Delta U is always zero.
However, that does not mean q = 0 and w = 0.
As I discussed in class for the isothermal expansion of an ideal gas: q = -w (meaning they cancel each other resulting in Delta U = 0)
Ask about this in my Monday review session.
However, that does not mean q = 0 and w = 0.
As I discussed in class for the isothermal expansion of an ideal gas: q = -w (meaning they cancel each other resulting in Delta U = 0)
Ask about this in my Monday review session.
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Re: delta U
Can someone also explain under what conditions is delta U equal to zero?
(Claire Woolson Dis 1K)
(Claire Woolson Dis 1K)
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Re: delta U
Recall that dU=q+w. For an isothermal expansion q=-w. Also, temperature does not change (hence isothermal), therefore q=0. If q=0 then w=0. Thus, dU=0+0=0.
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Re: delta U
In lecture yesterday for one of the review problems that asked to calculate deltaS of an isothermal reaction, Lavelle mentioned that U = (3/2)nRT. Thus, If there’s no change in temperature, since n, R, and T are all constant, that means there’s no change in internal energy.
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Re: delta U
With isothermal reversible expansion, work of expansion is being done by the system so energy is being lost. However, a heat reservoir also adds heat back into the system, so temperature remains constant. This is why q=-w.
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