Calculation of the change in internal energy of water

[NoahCygnus]

Homework Statement

If water vapor is assumed to be a perfect gas, molar enthalpy change for vaporization of 1 mol of water at 1 bar and 100 C is 41 kJ/mol . Calculate the internal energy, when 1 mol of water is vaporized at 1 bar pressure and 100 C.

Homework Equations

$$\Delta U = \Delta H = \Delta n_{g}RT$$

The Attempt at a Solution

The chemical equation is;
$$H_{2}O_{l} \longrightarrow H_{2}O_{g}$$
Then change in internal energy is related to change in enthalpy by;
$$\Delta U = \Delta H - \Delta n_{g}RT$$
Putting in the values;
$$\Delta U = 41 - (0)RT = 41 kJ/mol$$

However in the book they have given, $$\Delta n_{g} = 1$$. If the number of moles of water in liquid and gaseous phase are same , why isn't $$\Delta n_{g} = 0$$. ?

 

[Bystander]

Do you know what "Δ" means/indicates?

 

[NoahCygnus]

Do you know what "Δ" means/indicates?

Change in something. $\Delta n_{g}$ represents in this situation the difference of moles of gaseous water and liquid water, it should be zero. But in the book for some reason , it is 1.

 

[Bystander]

difference of moles of gaseous water and liquid water, it should be zero[insert] between initial/starting condition(s) and final condition(s) [/insert]. But in the book for some reason , it is 1.

Should it? Is it? What is the starting number?

 

[NoahCygnus]

Should it? Is it? What is the starting number?

In the chemical equation $1H_{2}O_{l} \longrightarrow 1H_{2}O_{g}$, 1 mol of water in liquid state is changing into 1 mol vapor so by that reasoning I came to the conclusion $\Delta n_{g}$ should be 0. Now I am truly lost... I checked a similar question, but there liquid water is being converted into ice and change in moles is $\Delta n_{g} = 0$. What is going on here.

 

[Bystander]

What is the initial number of moles of gas/vapor? What is the final number?

 

[Borek]

You are mistaking "change in number of moles of the substance" with "change in amount of the gas". In your equation number of moles stays the same, what about amount of gas?

 

[NoahCygnus]

What is the initial number of moles of gas/vapor? What is the final number?

Initially, there are zero number of moles of the vapor and the final number is 1. I thought $\Delta n_{g}$ gives a change in the number of moles of products and reactants.

 

[NoahCygnus]

You are mistaking "change in number of moles of the substance" with "change in amount of the gas". In your equation number of moles stays the same, what about amount of gas?

$\Delta n_{g} = n_{products} - n_{reactants}$ , the change relates products and reactants, it says $\Delta n_{g}$ is the difference between the moles of the right side of the chemical equation and the left side, so how am I making a mistake?

 

[Borek]

How much gas on the reactants side? How much gas on the products side?

 

[NoahCygnus]

How much gas on the reactants side? How much gas on the products side?

0 moles on the reactants side, 1 mole on the product side.

 

[Bystander]

What do the SUBSCRIPTS say?

 

[NoahCygnus]

What do the SUBSCRIPTS say?

It talks about the difference in the number of moles of gas between the products and reactants. As in the productside there are 0 moles of water vapour and on the right hand side there is 1 mole , $\Delta n_{g} = 1$, so I was confused I thought $\Delta n_{g}$ related to any state of matter on the reactants and products side.

 

[Chestermiller]

It is supposed to be the volume of one mole of water vapor minus the volume of one mole of liquid water (times the pressure). In your equation, the volume of one mole of liquid water is neglected compared to that of one mole of water vapor. So, from the ideal gas law, the pressure times the volume difference is $PV_g=RT$, where Vg is the volume of one mole of vapor.

 

[Bystander]

Left side:initial state in whatever state of aggregation::right side:final state in whatever state of aggregation.

In the chemical equation 1H2Ol⟶1H2Og1H2Ol⟶1H2Og1H_{2}O_{l} \longrightarrow 1H_{2}O_{g}, 1 mol of water in liquid state is changing into 1 mol vapor

You have got to pay attention to the subscripts. n_g = 0 in the initial state; n_g = 1 in the final state.

 

[NoahCygnus]

It talks about the difference in the number of moles of gas between the products and reactants. As in the productside there are 0 moles of water vapour and on the right hand side there is 1 mole , $\Delta n_{g} = -1$, so I was confused I thought $\Delta n_{g}$ related to any state of matter on the reactants and products side.

Left side:initial state in whatever state of aggregation::right side:final state in whatever state of aggregation.
You have got to pay attention to the subscripts. n_g = 0 in the initial state; n_g = 1 in the final state.

Left side:initial state in whatever state of aggregation::right side:final state in whatever state of aggregation.
You have got to pay attention to the subscripts. n_g = 0 in the initial state; n_g = 1 in the final state.

If a hypothetical reaction is taking place where $X_{s} + X_{s} \longrightarrow X_{2_{gas}}$ then $\Delta n$ will be again 1.
How about this one , $X_{s} + Y_{s} \longrightarrow XY_{l}$ , in this case $\Delta n$ should be 0. Am I getting it right ?

 

[Bystander]

Am I getting it right ?

In a word, "No." Just got back from an "excursion."

 

[Charles Link]

@NoahCygnus See @Chestermiller post 14. Also note that $H=U+PV$, so from that, given the $\Delta H$, enthalphy change that they supply you with, you can compute a $\Delta U$. The $PV$ term, in the gaseous state, as was pointed out in post 14, is computed as $PV=nRT$ for an ideal gas. The temperature for this equation is in degrees Kelvin. Meanwhile, you should have the value of the universal gas constant constant memorized. $R=1.987$ cal/(mole degree K) . I'll let you convert it to joules. $\\$ Note: In the liquid state, the $PV$ term is minimal, since the volume is minimal in the liquid state, and the contribution for the liquid to the $PV$ term in this computation can normally be ignored. $\\$ Since your background is somewhat limited according to your profile, I will write out the process in a little more detail: $H_{gas}=U_{gas}+(PV)_{gas}$, and $H_{liquid}=U_{liquid}+(PV)_{liquid}$. Meanwhile enthalpy change $\Delta H=H_{gas}-H_{liquid}$. To compute the change in internal energy $\Delta U=U_{gas}-U_{liquid}$, it is simply a little algebra. $\\$ (And it should be noted, the first equation that you wrote in the OP, that $\Delta H=\Delta U$, is incorrect).

 

[Chestermiller]

(And it should be noted, the first equation that you wrote in the OP, that $\Delta H=\Delta U$, is incorrect).

I think he meant to have a minus sign rather than the 2nd equal sign.

 

[Borek]

If a hypothetical reaction is taking place where $X_{s} + X_{s} \longrightarrow X_{2_{gas}}$ then $\Delta n$ will be again 1.
How about this one , $X_{s} + Y_{s} \longrightarrow XY_{l}$ , in this case $\Delta n$ should be 0. Am I getting it right ?

Beware: do you mean $\Delta n$ or $\Delta n_g$? (Your answers are correct if the latter).