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?
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.
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.
it really worked and helped me I really appreciate.
Well, to solve for w (work), you can use the equation:
w = -q
Since the process is isothermal, the temperature remains constant, and therefore there is no change in internal energy (∆U = 0). So, we can rewrite the equation as:
0 = q + w
Since we want to find both w and q, we need another equation to solve for q. In this case, we can use the equation for the enthalpy change (∆H):
∆H = q
So, we can substitute this into the equation above:
0 = ∆H + w
Now we have two equations:
w = -q
0 = ∆H + w
Using these equations, we can solve for w and q.
Now, since we already know that ∆H is -40.656 kJ, we can substitute this value:
0 = -40.656 + w
Solving for w:
w = 40.656 kJ
And since w = -q:
q = -40.656 kJ
So, the answers are:
w = 40.656 kJ (positive because work is done on the system)
q = -40.656 kJ (negative because heat is released from the system)
To solve for w (work) and q (heat), you need to use the first law of thermodynamics:
∆U = q + w
Given that ∆U = -37.55 kJ, you can plug this value into the equation:
-37.55 kJ = q + w
To solve for q, you need to know the value of w or vice versa. In this problem, since the process is isothermal (constant temperature), you can use the formula for work in an isothermal process:
w = -nRT * ln(V2/V1)
where n is the number of moles, R is the ideal gas constant, T is the temperature in Kelvin, V1 is the initial volume, and V2 is the final volume.
In this case, n = 1.00 mol, V1 refers to the volume of the gaseous H2O, and V2 refers to the volume of the liquid water. Since the sample is condensed from gas to liquid at constant volume, V2 = V1.
Substituting V2 = V1 into the work equation, you get:
w = -nRT * ln(1)
Since ln(1) = 0, the work done during this process is zero:
w = 0 kJ
Now you can solve for q:
-37.55 kJ = q + 0 kJ
q = -37.55 kJ
Therefore, the work done (w) is 0 kJ, and the heat transfer (q) is -37.55 kJ.