You should be able to graph data. Especially data that you just "make up". Let's make up some data for the work of a reversible vs. irreversible process. a. For a reversible process, work is: ∆w=−nRT∙ln( Vf/Vi) You have 0.25 mole of gas at a temperature of 298 K at an initial volume of 6.113 L. Please make a graph of work vs. final volume, which spans say 1 L to 20 L in 0.1 L increments. b. On the same graph, plot the irreversible work: ∆w = −Pext • (Vf - Vi) . Note that the external pressure must match the final volume. For example, if the final volume is 20 L, then Pext must be 31 kPa for PV=nRT to work. Thus, plot: ∆w = - (nRT/ Vf) • (Vf - Vi) where the final volume spans say 1L to 20 L in 0.1 L increments on the same graph in pt. a.
You should be able to graph data. Especially data that you just "make up". Let's make up some data for the work of a reversible vs. irreversible process.
a. For a reversible process, work is:
∆w=−nRT∙ln( Vf/Vi)
You have 0.25 mole of gas at a temperature of 298 K at an initial volume of 6.113 L. Please make a graph of work vs. final volume, which spans say 1 L to 20 L in 0.1 L increments.
b. On the same graph, plot the irreversible work: ∆w = −Pext • (Vf - Vi) . Note that the external pressure must match the final volume. For example, if the final volume is 20 L, then Pext must be 31 kPa for PV=nRT to work. Thus, plot:
∆w = - (nRT/ Vf) • (Vf - Vi)
where the final volume spans say 1L to 20 L in 0.1 L increments on the same graph in pt. a.
To plot a graph of reversible work done and irreversible work done for a isothermal process versus the final volume of the gas.
we are provided with of a gas with initial volume equals to and the temperature equals to .
The final volume varies from to 20 L with a span of 0.1 L increments.
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