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Isothermal expansion: reversible vs irreversible

  • Thread starter chemboy101
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  • #1
1. A sample of methane of mass 4.5 g occupies 12.7 L at 310 K. Assume that the gas
behaves ideally. (a) Calculate the work done when the gas expands isothermally against a
constant external pressure of 30.0 kPa until its volume has increased by 3.3 L. (b) Calculate the work that would be done if the same expansion occurred isothermally and reversibly




3. I have calulated (a) correctly I think by using w = -P dV to find w = -30 x 3.3 = 99 J.

However for (b) it states that the reaction is reversible. How does this change the equation for work calcuation?

Many thanks
CB
 

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  • #2
Redbelly98
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Welcome to Physics Forums.

(b) also states that the expansion is isothermal, whereas for (a) it was isobaric (constant pressure). For the isothermal expansion, the pressure will drop during the process.
 
  • #3
Andrew Mason
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1. A sample of methane of mass 4.5 g occupies 12.7 L at 310 K. Assume that the gas
behaves ideally. (a) Calculate the work done when the gas expands isothermally against a constant external pressure of 30.0 kPa until its volume has increased by 3.3 L.
PV=nRT. If a gas expands at constant pressure, the product PV increases so T has to increase (or you have to add more gas, which is not the case here). So it cannot be both isothermal AND isobaric.

So there is a significant initial pressure difference between the gas and the surroundings which decreases as the gas expands. The process is dynamic rather than quasi-static.

In the quasi-static, reversible case, the external pressure is the same as (ie infinitessimally less than) the pressure in the gas at all times. So you will have to do a bit of calculus for part b).

AM
 
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