Gas is isothermally and reversibly condensed

[vaazu]

Homework Statement

gas is isothermally and reversibly condensed
n=2(mol)
T=337K
H(vap)=35.3kJ/mol
find w, q, U, H

Homework Equations

The Attempt at a Solution

H=-H(vap)*2=-70.6kJ/mol
H=U+n(g)RT
w=-n(g)RT n(g)=-2 w=5.6kJ
So U=H-w U Should be -70.6-5.6=-76.2, but the book gives -65, Where am I wrong?
And why is H=q in this process?

 

[Ygggdrasil]

In the condensation from a gas to a liquid, what is the sign of ΔV? What then, should be the sign on w? Is work being done on the gas by the surroundings or is work being done on the surroundings by the gas?

Also, to answer your second question, ΔH = q for any reversible process done at constant pressure.
proof: From the definition of enthalpy, we know that
ΔH = ΔU + Δ(PV)
Next, we combine this expression with the first law of thermodynamics, ΔU = q + w, to get:
ΔH = q + w + Δ(PV)
At constant pressure, w = -integral(PdV) = -PΔV, and Δ(PV) = PΔV. Plugging these two expressions into the expression above gives the desired result:
ΔH = q

 

[Blueshift5]

I am not sure what specific system or process you are referring to in this question. However, I will provide a general response based on the information given.

First, let's define some terms:

- Isothermal: This means that the temperature (T) of the system remains constant throughout the process.
- Reversible: This means that the process can be reversed without any energy loss, and the system remains in thermodynamic equilibrium at all times.
- Condensed: This means that a gas is converted into a liquid or solid state.

Now, let's look at the equations you have provided:

H = U + n(g)RT: This is the equation for the enthalpy (H) of a system, which is the sum of its internal energy (U) and the product of the number of moles of gas (n(g)), the gas constant (R), and the temperature (T).

w = -n(g)RT: This is the equation for the work (w) done by a gas during an isothermal process, where n(g) is the number of moles of gas, R is the gas constant, and T is the temperature.

q = -w = n(g)RT: This is the equation for the heat (q) transferred during an isothermal process, where n(g) is the number of moles of gas, R is the gas constant, and T is the temperature.

Now, let's apply these equations to the given information:

- H(vap) = 35.3 kJ/mol: This is the enthalpy of vaporization, which is the amount of energy needed to convert 1 mole of a liquid into a gas at a constant temperature.
- n = 2 mol: This is the number of moles of gas.
- T = 337 K: This is the constant temperature of the system.

Using the equation H = U + n(g)RT, we can calculate the internal energy of the system:

H = U + n(g)RT
35.3 kJ/mol = U + (2 mol)(8.314 J/mol*K)(337 K)
U = -70.6 kJ/mol

Note that the units of R must be in Joules per mole Kelvin (J/mol*K) to get the correct units for U.

Next, we can calculate the work done by the gas using the equation w = -n(g)RT:

w =