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Differentials | Example 2

The example of differentials is the following

Dulong’s law states that if $P$ atmospheres is the absolute pressure of a saturated vapor at a temperature of $T$ Celsius, then $P = \left( \frac{40 + T}{140} \right)^{5}$, where $T>80$. Calculate the instantaneous variation rate of $P$ with respect to $T$, when a) $T = 240$ b) $P = 1$

To start solving the problem and find the instantaneous variation rate of $P$ with respect to $T$ when $T=240$ and then when $P=1$, first we will find the derivative of $P$, remember that it is the derivative of $P$ with respect to $T$:

$$P = \left( \cfrac{40 + T}{140} \right)^{5} $$

The derivative of the equation $ P $ is calculated with the chain rule that will make us pass the $5$ to multiply and now the exponent will be $5-1=4$ but the expression inside the parentheses is conserved. Now, this expression that we mentioned has to be multiplied by the derivative of what is inside the parentheses, which can easily be seen with the naked eye:

$$\cfrac{dP}{dT} = 5 \left( \cfrac{40 + T}{140} \right)^{4}\left( \cfrac{40 + T}{140}\right)’$$

The derivative of the parenthesis can be expressed as a sum of derivatives, where $\frac{40}{140}$ equals zero and $\frac{T}{140}$ equals $\frac{1}{140}$:

$$\cfrac{dP}{dT} = 5\left( \cfrac{40+T}{140} \right)^{4} \left(\cfrac{1}{140} \right)$$

We simplify the $5$ and the $140$:

$$\cfrac{dP}{dT} = \cfrac{1}{28} \left( \cfrac{40+T}{140}\right)^{4}$$

Let’s go with part a) when $T=240$

This subsection is simple, we will only take the differential that we calculated and in $T$ we will substitute the value of $240$:

$$\cfrac{dP}{dT} = \cfrac{1}{28} \left( \cfrac{40 + T}{140} \right) ^{4} = \cfrac{1}{28} \left( \cfrac{40 + 240}{140} \right)^{4}$$

Solving the sum and simplifying the fraction, we will have the following:

$$\cfrac{dP}{dT} = \cfrac{1}{28} \left( \cfrac{280}{140} \right)^{4} = \cfrac{1}{28} \left( 2 \right)^{4}$$

We will make the exponent and then multiply by the fraction of $\frac{1}{28}$:

$$\cfrac{dP}{dT} = \cfrac{1}{28} \left(16 \right) = \cfrac{4}{7}$$

Thus we will obtain that our instantaneous rate of variation of $P$ with respect to $T$, when $T=240$ is:

$$\cfrac{dP}{dT} = \cfrac{4}{7} \ atm / ^{\text{o}}\text{C}$$

It is necessary to calculate subsection b) of the instantaneous variation rate of $P$ with respect to $T$ when $P=1$

In order to solve this exercise, we first need to know how much $T$ is worth when $P=1$, for that we take our equation $P$ and substitute $P=1$:

$$P = \left( \cfrac{40 + T}{140} \right) ^{5} \ \Rightarrow \ 1 = \left( \cfrac{40 + T}{140} \right)^{5}$$

Eliminate the exponent to the fifth by taking root to the fifth to all the equality, and the root fifth of $1$ is equal to $1$:

$$\sqrt[5]{1} = \sqrt[5]{\left( \cfrac{40 + T}{140} \right)^{5}}$$

$$1 = \cfrac{40 + T}{140}$$

Now we spend the $140$ multiplying to $1$:

$$140 = 40 + T$$

And the $40$ will subtract the $140$ so that we get the value of $T$ when $P=1$:

$$T = 100$$

Now we can find the instantaneous rate of variation of $P$ with respect to $T$ when $P=1$, take our calculated derivative and substitute the value of $T$ with $100$:

$$\cfrac{dP}{dT} = \cfrac{1}{28} \left( \cfrac{40 + T}{140} \right)^{4}$$

$$\cfrac{dP}{dT} = \cfrac{1}{28} \left( \cfrac{40 + 100}{140} \right)^{4}$$

Let’s make the sum of the fraction and simplify the same fraction:

$$\cfrac{dP}{dT} = \cfrac{1}{28} \left(\cfrac{140}{140} \right)^{4} = \cfrac{1}{28} \left( 1 \right)^{4}$$

Raising the $1$ to the fourth power and multiplying it by the $\frac{1}{28}$, we get the following:

$$\cfrac{dP}{dT} = \cfrac{1}{28}$$

And so we finally get that our instantaneous rate of variation of $P$ with respect to $T$ when $P=1$ is:

$$\cfrac{dP}{dT} = \cfrac{1}{28} \ atm/^{\text{o}}\text{C}$$

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