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TheEyeOfInnos

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- Homework Statement
- An ideal gas goes through a transformation that in (pV) coordinates is a straight line. In the final state we have ##V_{2}=4V_{1}## and ##p_{2}=p_{1}/3##. We are asked to calculate the ratio between the change in internal energy during heating and the change in internal energy during cooling.

- Relevant Equations
- ##p=a-bV##

##pV = nRT##

##pV^{\gamma}=const.##

##\Delta U = nC_{V}\Delta T##

A linear transformation is described by:

$$p=a-bV$$

From that we can find $a$ and $b$:

$$b=\frac{p_{2}-p_{1}}{V_{1}-V_{2}} = \frac{2}{9}\frac{p_{1}}{V_{1}}$$

$$a=p_{1}+bV_{1}=\frac{11}{9}p_{1}$$

I considered an adiabatic process that intersects the linear transformation to find the point up until the system received heat:

$$pV^{\gamma}=\text{const.}$$

$$\left(a-bV\right)V^{\gamma}=\text{const.}$$

Differentiating wrt to volume:

$$V_{P} = \frac{\gamma}{\gamma+1}\frac{a}{b}=\frac{\gamma}{\gamma+1}\frac{11}{2}V_{1}$$

now we can obtain:

$$p_{P} = a-bV_{P}=\frac{22}{18}\frac{\gamma}{\gamma+1}p_{1}$$

In order to find the temperature ##T_{P}## I used ##p_{1}V_{1}=nRT_{1}## and ##p_{P}V_{P}=nRT_{P}##

$$T_{P} = T_{1}\cdot p_{P}\cdot V_{P} = \frac{121}{18}\frac{\gamma}{\gamma+1}T_{1}$$

In order to find the temperature ##T_{2}## I used ##p_{1}V_{1}=nRT_{1}## and ##p_{2}V_{2}=nRT_{2}##

$$T_{2}=T_{1}\cdot p_{2}\cdot V_{2}=\frac{4}{3}T_{1}$$

Now:

$$\frac{\Delta U_{heating}}{\Delta U_{cooling}} = \frac{T_{P}-T_{1}}{T_{2}-T_{P}}$$

But I cannot get rig of ##\gamma## and the problem does not specify which type of gas we are dealing with. The solution apparently is -1.96.

$$p=a-bV$$

From that we can find $a$ and $b$:

$$b=\frac{p_{2}-p_{1}}{V_{1}-V_{2}} = \frac{2}{9}\frac{p_{1}}{V_{1}}$$

$$a=p_{1}+bV_{1}=\frac{11}{9}p_{1}$$

I considered an adiabatic process that intersects the linear transformation to find the point up until the system received heat:

$$pV^{\gamma}=\text{const.}$$

$$\left(a-bV\right)V^{\gamma}=\text{const.}$$

Differentiating wrt to volume:

$$V_{P} = \frac{\gamma}{\gamma+1}\frac{a}{b}=\frac{\gamma}{\gamma+1}\frac{11}{2}V_{1}$$

now we can obtain:

$$p_{P} = a-bV_{P}=\frac{22}{18}\frac{\gamma}{\gamma+1}p_{1}$$

In order to find the temperature ##T_{P}## I used ##p_{1}V_{1}=nRT_{1}## and ##p_{P}V_{P}=nRT_{P}##

$$T_{P} = T_{1}\cdot p_{P}\cdot V_{P} = \frac{121}{18}\frac{\gamma}{\gamma+1}T_{1}$$

In order to find the temperature ##T_{2}## I used ##p_{1}V_{1}=nRT_{1}## and ##p_{2}V_{2}=nRT_{2}##

$$T_{2}=T_{1}\cdot p_{2}\cdot V_{2}=\frac{4}{3}T_{1}$$

Now:

$$\frac{\Delta U_{heating}}{\Delta U_{cooling}} = \frac{T_{P}-T_{1}}{T_{2}-T_{P}}$$

But I cannot get rig of ##\gamma## and the problem does not specify which type of gas we are dealing with. The solution apparently is -1.96.

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