- #1

lee403

- 16

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**1. For a van der Waals gas π**

_{T}=a/V_{m}^{2}. Calculate ΔU_{m}for the isothermal expansion of nitrogen gas from an initial volume of 1.00 dm^{3}to 24.8 dm^{3}at 298K. What are values for q and w?## Homework Equations

I looked up a for N

_{2}gas to be 1.352 atm*dm

^{6}/mol

^{2}

dU(V,T)=(∂U/∂V)

_{T}dV+(∂U/∂T)

_{V}dT

(∂U/∂V)

_{T}=π

_{T}

(∂U/∂T)

_{V}=C

_{v}

3. The Attempt at a Solution

3. The Attempt at a Solution

I start with (∂U/∂V)

_{T}=π

_{T}and substitute in π

_{T}and C

_{v}

which gives dU(V,T)=π

_{T}dV+C

_{v}dT

Since this is isothermal dt=0 so dU(V,T)=π

_{T}dV

I integrate both sides ∫dU=∫π

_{T}dV. π

_{T}= a/V

_{m}

^{2}.

a is a constant so it comes out of the integral leaving

ΔU=a∫dV/V

_{m}

^{2}.= a*n

^{2}∫ dV/V

^{2}

Evaluating the integral

ΔU= -((a*n

^{2})/2)*V

^{-3}.

After plugging in values for a, V

_{i}, and V

_{f}I get:

ΔU=-((1.352 *n

^{2})/2)*[(1/24.8

^{3})-(1/1

^{3})]

I get 0.676 *n

^{2}J/mol

^{2}. I just don't know what to do with the moles.

I know how to get work which is just w=-nrt∫dV/V and the solution is -7.95 kJ of work

but without ΔU I cannot calculate q.