The $\Delta V$ is supposed to be the volume of the vapor minus the volume of the liquid. The volume of the liquid is only 0.025 liters, which they neglect.

I don't know where they get the 12470.6 from. I'm guessing this is an error.

(0.082)(373)(25)/18=42.48. You solve it basically they way you said.

Δ(PV)=42.48 liter−atm =1029 cal=42.48 liter-at = 1029 cal

So, ΔU=13500−1029=12471 cal

The $\Delta V$ is supposed to be the volume of the vapor minus the volume of the liquid. The volume of the liquid is only 0.025 liters, which they neglect.

I don't know where they get the 12470.6 from. I'm guessing this is an error.

(0.082)(373)(25)/18=42.48. You solve it basically the way you said.

Δ(PV)=42.48 liter−atm =1029 cal=42.48 liter-at = 1029 cal

So, ΔU=13500−1029=12471 cal

The $\Delta V$ is supposed to be the volume of the vapor minus the volume of the liquid. The volume of the liquid is only 0.025 liters, which they neglect.

I don't know where they get the 12470.6 from. I'm guessing this is an error.

The $\Delta V$ is supposed to be the volume of the vapor minus the volume of the liquid. The volume of the liquid is only 0.025 liters, which they neglect.

I don't know where they get the 12470.6 from. I'm guessing this is an error.

(0.082)(373)(25)/18=42.48. You solve it basically they way you said.

Δ(PV)=42.48 liter−atm =1029 cal=42.48 liter-at = 1029 cal

So, ΔU=13500−1029=12471 cal