Isobaric/isochoric (?) heating of an ideal gas

[MexChemE]

Homework Statement

Consider an ideal gas with $C_V=6.76 \frac{cal}{mol \cdot K}$. Calculate $\Delta H$ and $\Delta U$ when ten moles of this gas are heated from 273.15 K to 373.15 K.

Homework Equations

$$\Delta H = \Delta U + P\Delta V$$
$$Q=n C_V \Delta T$$

The Attempt at a Solution

As I'm given the heat capacity at constant volume I'm assuming this is an isochoric process. That means $W=0$. Therefore, $\Delta U=Q$.
$$\Delta U=Q= (10 \ mol) \left(6.76 \frac{cal}{mol \cdot K} \right) (100 \ K) = 6760 \ cal$$
Now, for the change in enthalpy we have:
$$\Delta H = 6760 \ cal + P\Delta V$$
This is where I'm having trouble. Should I cancel the second term in the above equation? And so have: $\Delta H=\Delta U$. But, if the gas is heated at constant volume pressure should increase, that means I should consider the PV term in the last equation. I wasn't provided with initial or final values for pressure and volume, so there's not enough information to use PV=nRT. What should I do?

 

[Juxtaroberto]

But the last term isn't V, it's ΔV, so no matter what P is, PΔV for an isochoric process will be 0, because volume doesn't change (ΔV=0). I feel that this is a safe assumption, since you were given C_V, which is the specific heat at constant volume.

 

[MexChemE]

Could the last term be Δ(PV)? I just realized I can get (PV)₁ and (PV)₂ from PV=nRT with the data I was provided.

 

[MexChemE]

Sorry to double post but I think I figured it out. The standard definition of the change of enthalpy: ΔH = ΔU + Δ(PV). The case ΔH = ΔU + PΔV is only for isobaric processes. The process here is isochoric, so we have no work done by the system, but we had a change in pressure. Therefore, ΔH = Q + PΔV + VΔP. I can cancel out the PΔV term, and I can get VΔP with the ideal gas law. So:
$$\Delta H = 6790 \ cal + (10 \ mol) \left(1.987 \ \frac{cal}{mol \cdot K} \right)(100 \ K) = 8777 \ cal$$

Is it right this time?

 

[ehild]

Nothing is said about the process of heating. The internal energy of n mol of an ideal gas is U=nCvT, so
$\Delta U=nCv \Delta T$
for any process. H is defined as H=U+PV. Substituting P=nRT/V from the ideal gaw law, H=nT(Cv+R) =nTCp.
$\Delta H=nCp \Delta T$.
Your method and result are correct.

ehild

 

[MexChemE]

Thank you!