The example is:
Engineers designing new piston engines and turbines need to understand how work and heat are involved in various compression and expansion cycles. Suppose that 1.00 mol of ideal gas molecules at an initial pressure of 3.00 atm and 292 K expands against a constant external pressure of 0.20 atm from 8.00 L to 20.00 L by two different paths. (a) Path A is an isothermal, reversible expansion. (b) Path B, a hypothetical alternative to path A, has two steps. In step 1, the gas is cooled at constant volume until its pressure has fallen to 1.20 atm. In step 2, it is heated and allowed to expand against a constant pressure of 1.20 atm until its volume is 20.00 L and cap t equals 292 cap k. Determine for each path the work done (w), the heat transferred (q), and the change in internal energy .
Why is q=-w in both parts (a) and (b) of the problem?
Example 4B.1
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Re: Example 4B.1
Postby Becky Belisle 1A » Tue Feb 12, 2019 9:18 am
Isothermal means the change in internal energy equals zero, and since Δ U = q + w, q = -w.
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Re: Example 4B.1
Postby Becky Belisle 1A » Tue Feb 12, 2019 9:21 am
For part b, internal energy is a state function, so although 2 different path are given, Δ U is still equal to zero. State functions are path independent.
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