A sample of 1.00 mol H2O(g) is condensed isothermally and reversibly to liquid water at 100°C. The standard enthalpy of vaporization of water at 100°C is 40.656 kJ mol−1. Find w, q, ∆U, and ∆H for this process.
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I can solve for ∆H: n * standard ∆H * -1 because reverse of vaporization.
-40.656 kJ
I can solve for ∆U:
∆H = ∆U + ∆n(gas)*RT
-40.656 = ∆U + -1 * -R * 373.15K
∆U = -37.55 kJ
Now, how do I figure out w and q
I know that
∆U = q + w
but beyond that, how do I solve?
3 answers
The pressure is constant, and so the heat at constant pressure is qp = - dH vaporization. W=-pdV and so since p=pext=1atm and dV = (vf - vi) = vliquid - vvapor, we'll say that since the volume of the vapor is significantly greater than that of the liquid, then vf-vi is approximately equal to -volume of vapor = - nRT/pext. Solve for these and then you get dU.
Well, the pressure here is not constant, but the temperature is. Reversible implies that there are infinitesimal changes in certain parameters.
For isothermal, reversible systems
delta H=0
delta U=0
q=-w
and w=-nRTln(Vf/Vi)
You can find your Vf by using the density constant for liquid water which s essentially 1.0 g/mL.
I used SATP laws for an ideal gas to find the volume of water at room temp. Then I converted to the volume at higher temp using the relation V1/T1=V2/T2. Hope this works.
For isothermal, reversible systems
delta H=0
delta U=0
q=-w
and w=-nRTln(Vf/Vi)
You can find your Vf by using the density constant for liquid water which s essentially 1.0 g/mL.
I used SATP laws for an ideal gas to find the volume of water at room temp. Then I converted to the volume at higher temp using the relation V1/T1=V2/T2. Hope this works.
it really worked and helped me I really appreciate.
1 answer