Work and ideal gas law
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Work and ideal gas law
Can someone explain the relationship between work and the ideal gas law? I believe a problem was done in class where work was found using the ideal gas law. Thank you in advance.
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Re: Work and ideal gas law
The ideal gas law is PV = nRT, and work is equal to -P(change in V). So, if you know the number of moles and the temperature of a reaction, you can solve for work using the idea that work = -nRT. I think there was an example of this in Friday's lecture.
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Re: Work and ideal gas law
In the example we did in class, we had an exothermic reaction with ΔH = -50kJ occuring in an open beaker producing net 8 moles of gas at 25C, and we had to calculate the change in internal energy. Since ΔU = ΔH - PΔV, we had to use the ideal gas law to calculate PΔV, setting it equal to ΔnRT since the volume and number of moles of gas changed. You would then get PΔV equal to 20 kJ and therefore ΔU = -50 - 20 = -70 kJ.
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Re: Work and ideal gas law
Work = -P ∆V when the external pressure is constant.
By the ideal gas law, PV = nRT, and P∆V = ∆(nRT)
So any work done under constant pressure = - ∆(nRT), where the change in volume can result from a change in the number of moles of gas (∆n) or a change in the temperature (∆T).
By the ideal gas law, PV = nRT, and P∆V = ∆(nRT)
So any work done under constant pressure = - ∆(nRT), where the change in volume can result from a change in the number of moles of gas (∆n) or a change in the temperature (∆T).
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