Isobaric Compression: Calculate Q, W and \Delta E

[SoggyBottoms]

Homework Statement

We have an ideal gas of N particles with mass m and temperature T and volume V.

a) Calculate $\langle E_{kin} \rangle$

We now reversibly compress the gas from volume V to V/2. During this compression heat Q is added, work W is done on the gas and the energy of the gas changes by $\Delta E$.

b) Calculate Q, W and $\Delta E$ in case the compression is isobaric.

The Attempt at a Solution

a) This one I know how to do, the answer is $\langle E_{kin} \rangle = \frac{3}{2}N k_B T$

b) The change is isobaric and the work done on the gas is positive, so $W = p \Delta V = p(V - \frac{V}{2}) = \frac{N k_B \Delta T}{2}$.

We also have that $\Delta E = \Delta U = C_V \Delta T = \left(\frac{\partial \langle E_{kin} \rangle}{\partial T}\right)_V \Delta T = \frac{N k_B \Delta T}{2}$.

Now: $$\Delta Q = \Delta U + \Delta W = \Delta T (\frac{N k_B}{2} + \frac{N k_B}{2}) \\<br /> = N k_B \Delta T$$

So $Q = N k_B T$

Is this correct?

 

[ehild]

a) This one I know how to do, the answer is $\langle E_{kin} \rangle = \frac{3}{2}N k_B T$

It is true when the gas is mono-atomic.

b) The change is isobaric and the work done on the gas is positive, so $W = p \Delta V = p(V - \frac{V}{2}) = \frac{N k_B \Delta T}{2}$.

It is NK_bT₁/2. Do not write ΔT.

We also have that $\Delta E = \Delta U = C_V \Delta T = \left(\frac{\partial \langle E_{kin} \rangle}{\partial T}\right)_V \Delta T = \frac{N k_B \Delta T}{2}$.

Cv=3/2 NK_b for the mono-atomic gas. What is the change of temperature during the isobaric compression?

ehild