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Chapter 6: Problem 8

and volume, \(V\), of a gas in a cylinder are related by \(P V^{n}=C\) (where \(C\) is a constant) Show that $$ \frac{\mathrm{d}^{2} P}{\mathrm{~d} V^{2}}=\frac{n(n+1) P}{V^{2}} $$

Short Answer

Expert verified
\(\frac{\mathrm{d}^2 P}{\mathrm{d} V^2} = \frac{n(n+1) P}{V^2} \)

Step by step solution

01

- Express Pressure as a Function of Volume

Starting from the given equation, express pressure, \(P\), as a function of volume, \(V\):\[P V^n = C\] Solving for \(P\), one gets: \[P = \frac{C}{V^n}\]
02

- Differentiate Pressure with Respect to Volume

Differentiate \(P = \frac{C}{V^n}\) with respect to \(V\):\[\frac{\mathrm{d} P}{\mathrm{d} V} = \frac{\mathrm{d}}{\mathrm{d} V} \left( \frac{C}{V^n} \right)\]Using the power rule for derivatives: \[\frac{\mathrm{d} P}{\mathrm{d} V} = C \cdot \frac{-n}{V^{n+1}}\]Since \(C = PV^n\), substitute \(PC = V^n\) back in the equation: \[\frac{\mathrm{d} P}{\mathrm{d} V} = -n \left( \frac{P}{V} \right)\]
03

- Differentiate Again with Respect to Volume

Differentiate \(\frac{\mathrm{d} P}{\mathrm{d} V} = -n \left( \frac{P}{V} \right)\) with respect to \(V\):\[\frac{\mathrm{d}^2 P}{\mathrm{d} V^2} = \frac{\mathrm{d}}{\mathrm{d} V} \left( -n \frac{P}{V} \right)\]Using the product rule for differentiation: \[\frac{\mathrm{d}^2 P}{\mathrm{d} V^2} = -n \left( \frac{\mathrm{d} P}{\mathrm{d} V} \cdot \frac{1}{V} + P \cdot \frac{-1}{V^2} \right)\]
04

- Substitute \( \frac{\mathrm{d} P}{\mathrm{d} V} \)

Substitute \(\frac{\mathrm{d} P}{\mathrm{d} V} = -n \left( \frac{P}{V} \right)\) back into the equation: \[\frac{\mathrm{d}^2 P}{\mathrm{d} V^2} = -n \left( -n \frac{P}{V^2} - \frac{P}{V^2} \right)\]Simplify: \[\frac{\mathrm{d}^2 P}{\mathrm{d} V^2} = -n \left( \frac{P}{V^2} (n+1) \right) \]This further simplifies to: \[\frac{\mathrm{d}^2 P}{\mathrm{d} V^2} = \frac{n(n+1) P}{V^2}\]
05

- Conclusion

We have shown that \(\frac{\mathrm{d}^2 P}{\mathrm{d} V^2} = \frac{n(n+1) P}{V^2} \)

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Pressure-Volume Relationship
To understand the pressure-volume relationship, we start with the given equation: \[P V^n = C\]. This equation shows that the product of the pressure (P) and the volume (V) raised to a power (n) is a constant (C).
It's a key concept in thermodynamics and relates to how pressure and volume change in various states of a gas.
By rearranging the equation, we can express pressure as: \[P = \frac{C}{V^n}\].
This simple manipulation sets the stage for further differentiation steps done in the solution.
Second Derivative
The second derivative in calculus helps us understand how the rate of change of a function's rate of change behaves.
To find it for our pressure-volume relationship, we first differentiate pressure with respect to volume once and then differentiate the resulting expression once more.
Using the equation \[P = \frac{C}{V^n}\], we apply the first derivative, giving \[ \frac{\text{d}P}{\text{d}V} = -n \frac{C}{V^{n+1}} \]. Simplifying and using the original equation, it turns into \[ \frac{\text{d}P}{\text{d}V} = -n \frac{P}{V} \].
The second derivative involves differentiating this new expression again: \[ \frac{\text{d}^2P}{\text{d}V^2} \]. Using differentiation rules helps in achieving this final formula: \[ \frac{\text{d}^2P}{\text{d}V^2} = \frac{n(n+1)P}{V^2} \].
This illustrates how pressure changes not just in response to volume but also how that rate of change itself adapts.
Product Rule
When differentiating a product of two functions, the product rule is essential.
The rule states: \[ \frac{\text{d}}{\text{d}x}(uv) = u \frac{\text{d}v}{\text{d}x} + v \frac{\text{d}u}{\text{d}x} \] where u and v are both functions of x.
In our problem, we face such a situation while differentiating \[ -n \frac{P}{V} \].
Applying the product rule, we get: \[ \frac{\text{d}}{\text{d}V}( -n \frac{P}{V}) = -n \frac{\text{d}P}{\text{d}V} \frac{1}{V} - n \frac{P}{V^2} \]. This step is crucial in simplifying and achieving the accurate form of the second derivative.
Understanding this concept ensures mastery in tackling more complex differentiation problems.

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Most popular questions from this chapter

Find \(\frac{\mathrm{d} y}{\mathrm{~d} x}\) for each of the following cases: a \(x^{2}+y^{2}-x y^{2}=5\) *b \(\cos \left(x^{2}+y\right)-2 x y=0\) c \(\ln (y)=\ln (x-1)-\ln \left(x^{2}\right)\) d \(\ln (x y)+2 y=0\) e \(e^{x+y}+\sin (x)=0\)

Differentiate each of the following expressions with respect to the variables used, simplify your result where possible: a \(t \sin (t)\) b \(x \ln (x)\) c \(\theta e^{\theta}$$\begin{array}{ll}\mathbf{d} e^{u} \sin (u) & \mathbf{e} \sin (\alpha) \cos (\alpha) \\ \mathbf{f} \frac{q}{1+q} & \mathbf{g} \frac{\cos (a)}{1+\sin (a)} \\ \boldsymbol{h} \frac{e^{z}}{\cos (z)} & \text { i } M^{3} \ln (M) \\ \mathbf{j} \frac{\beta^{2}}{\sin (\beta)} & { }^{*} \mathbf{k} \frac{h}{\sqrt{1+h^{2}}} \\ \mathbf{I} J \ln [\sin (J)] & \mathbf{m} e^{\Sigma} \sin (2 \Sigma) \\ \text { n } \sqrt{\frac{Z+1}{Z-1}} & { }^{*} \mathbf{o} \Sigma e^{\Sigma} \sin (5 \Sigma) \\ * \mathbf{p} \ln \left(\frac{\delta^{2}+1}{\delta^{2}-1}\right) & \end{array}\)

In the following, \(x\) represents the displacement of a particle in a horizontal line at time \(t\). In each case find expressions for the velocity, \(v=\frac{\mathrm{d} x}{\mathrm{~d} t}\), and acceleration, \(a=\frac{\mathrm{d}^{2} x}{\mathrm{~d} t^{2}}\) at time \(t\) : a \(\quad x=t^{3}-10 t^{2}+40 t-30\) b \(x=7 t^{3}-50 t^{2}+50 t\) c \(\quad x=5 \cos (2 t)\) d \(\quad x=3 \cos (10 \pi t)+5 \sin (10 \pi t)\)

i Plot the function \(y=10 x^{6}\) for \(x=0\) to 3 ii Plot on different axes the function \(y=10 x^{6}\) for \(x=1.4\) to \(1.6\) \(x=1.49\) to \(1.51\) and \(x=1.499\) to \(1.501\). iii Evaluate \(\lim _{h \rightarrow 0} \frac{10(1.5+h)^{6}-10(1.5)^{6}}{h}\). This is \(\frac{\mathrm{d} y}{\mathrm{~d} x}\) of \(y=10 x^{6}\) at \(x=1.5\).

[structures] The bending moment, \(M(\mathrm{kN} \mathrm{m})\), of a beam of length \(8 \mathrm{~m}\) is given by $$ M= \begin{cases}2.16 x & 0 \leq x<5 \\ 28.8-3.6 x & 5 \leq x \leq 8\end{cases} $$ where \(x\) is the distance along the beam. The shear force, \(V\), is given by \(V=\frac{\mathrm{d} M}{\mathrm{~d} x}\). Sketch \(V\) for \(0 \leq x \leq 8\)
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