This is a proof of the formula used to calculate initial/final pressures and volumes during an adiabatic process once the process’ pressure/volume trajectory is known.

During an adiabatic process no heat enters or leaves the system which implies that the work done by the system is equal to the negative of the change in internal energy of the system. To find the work done by the system, we first start with the ideal gas formula because it relates the changing pressure and changing volume of the system to the changing temperature of the system. For an adiabatic process, this formula looks like $Vdp + Pdv = nRdt$, where $n$ is the amount of the gas usually measured in moles, and R is the ideal gas constant. After dividing the left side of the equation by $PV$ and the right side by $nRdT$ we obtain the differential equation $\frac{1}{P} dp + \frac{1}{V} dv = \frac{1}{T} dt$. Now, we use the differential form of the First Law of Thermodynamics Formula to obtain a worthy substitution for $dt$ and $\frac{1}{T}$. Because no heat enters or leaves the system of an adiabatic process, the differential equation of this formula becomes $du = -dw$. Now we can substitute $Pdv$ for $dw$ because our adiabatic system changes volume by an extremely small amount, allowing us to regard the pressure on the gas as being constant throughout the process. This leaves us with $du =-Pdv$. Now if we take, for example some other isometric process with initial and final temperatures, we will get the same change in internal energy or $du$. Because no work is done during an isometric process, the change in internal energy for  the isometric process is equivalent to the change in heat for the process or $dq$. After substituting $nC_{v}dt$ for $dq$, we can then substitute this in for $du$ in the equation of our adiabatic process which then gives us $nC_{v}dt$. We can now solve for $dt$ and then substitute that expression into $dt$ in our initial differential equation as a constraint. Now, to find a suitable substitution for $\frac{1}{T}$, we go to the ideal gas equation, and solve for $\frac{1}{T}$. We then plug this into the initial differential equation to obtain the expression $\frac{1}{P} dp + \frac{1}{V} dv = - \frac{R}{C_{v}V} dv$. Now we can set the equation equal to zero which gives us $\frac{1}{P}dp + (1+\frac{R}{C_{v}})\frac{1}{V} dv =0$. The term $(1+\frac{R}{C_{v}})$ can be written as $\frac {C_{v}}{C_{v}} + \frac{R}{C_{v}} = \frac{C_{v} +R}{C_{v}} = \frac{C_{p}}{C_{v}}= \gamma$. Now, we can integrate both sides of the equation to obtain $\int{\frac{1}{P}} dp + \int{\gamma\frac{1}{V}} dv = \int{0}$. After integrating, we obtain the equation $lnP + \gamma lnV = constant$. We can then exponentiate both sides to obtain $e^{lnP}e^{lnV^{\gamma}} = constant \Rightarrow PV^{\gamma} = constant$.

Now to derive the formula for the work done by a gas during an adiabatic process, we will begin with the equation $W = \int{P}dv$. To find a worthy substitute for $P$, we will turn to the equation just derived above, and apply it to an initial point and a final point. We will then solve for $P$, and substitute it for $P$ in the integral above which gives us $\int{P_{i}V_{i}^{\gamma}V^{-\gamma}}$. Then we can pull $P_{i}$ and $V_{i}^{\gamma}$ out of the integral and integrate what is left from $V_{i}$ to $V_{f}$. This leaves us with $\frac{P_{i}V_{i}^{\gamma}}{1-\gamma}[V_{f}^{1-\gamma}-V_{i}^{1-\gamma}]$. Now we can use the fact that $P_{i}V_{i}^{\gamma}= P_{f}V_{f}^{\gamma}$ to substitute $P_{f}V_{f}^{\gamma}$ for $P_{i}V_{i}^{\gamma}$ in our equation , which leads us to the final equation for the work done by a gas during an adiabatic process. The final equation is $W = \frac{P_{f}V_{f} - P_{i}V_{i}}{i-\gamma}$

In the figure below a box (total mass $m_1$ = 1.65 kg) and another box (total mass $m_2$ =3.30 kg) slide down an inclined plane while attached b a massless rod parallel to the plane. The angle if incline is $\Theta$ = 30.0°. The coefficient of kinetic friction between $m_1$ and the incline is $\Theta$ =30.0°. The coefficient of kinetic friction between $\mu_1$ and the incline is $\mu_1$ = .226; that between $m_2$ and the incline is $\mu_2$ = .113. Compute the tension in the rod and the magnitude of the common acceleration of the two boxes.

Once the force vectors are labeled and the axis components are resolved we can set up Force Systems to compute the tension and magnitude of common acceleration. The systems are as follows:

System 1a: m1, perpendicular, a = 0

Force balance, $|C_1| = |m_1gcos \theta|$.

System 1b: m1, parallel, a

$+|m_1gsin \theta|+|T|-|F_{f1}|=m_1a$

System 2a: m2, perpendicular, a = 0

Force balance, $|C_2| = |m_{2}gcos \theta|$

System 2b: m2, parallel, a

$+|m_2gsin \theta| - |T| - |F_{f2}|=m_2a$

$F_{f1}$ Model

$|F_{f1}| = .226C_1$

$F_{f2}$ Model

$|F_{f2}| = .113C_2$

Since there is no acceleration in the vertical dimension we only add the horizontal systems for both masses to obtain $\sum F$ of the system. This gives us:

$gsin \theta(m_1+m_2) - F_{f1} - F_{f2} = (m_1 + m_2)a$

$a = g[sin 30(m_1+m_2) - (\mu m_1 + \mu m_2)cos 30$

$a = 3.62m/s^2$

$|T| = |m_1a| + |F_f|-|m_1gsin \theta |$

$|T| = |(1.65)(3.62)| + |.726m_1gcos30|-|(1.65)(9.8)sin30|$

$|T| = 1.05 N$

# Tension Problem

The weight of a crate is 566 newtons. Find the tension in each of the supporting cables shown in the figure.

The Newton is the unit used for Force and is equivalent to the force required to give a mass of 1 kg an acceleration of 1 ms^-2. We are going to calculate the tension force in newtons that exists in each supporting cable. We can begin this problem by imagining the weight of the box as a force exerted downward and coinciding with the z-axis. This can be written in component form as W=<0,0,-115>. The end point where all of the cables connect occurs at A=(0,0,-115). If we take the point, (0,0,0), at the origin we can imagine it to exist on the horizontal plane which represents the surface that the cables are suspended from. We can then subtract the coordinates of A from this point to yield a weight vector in component form. Next we can create coordinate points which represent where each cable is connected to the horizontal plane. This is done using their distances from the x and y axis, and yields the three points: B-(0,70,0), C-(-60, 0, 0), and D-(45, -65, 0). Next, we can subtract the x, y, and z coordinates of A from each coordinate of each connecting point to obtain the line segments AB=<0,70,115>, AC=<-60, 0, 115>, and AD=<45, -65, 115>. Each are in component form so that they are dissociated from a particular place in space as is the weight force vector.

Now we can divide AB by the magnitude of ||AB|| which will yield a unit vector in the direction of B. Then all that is missing from the unit vector is the magnitude which gives us the length of the supporting cable. After repeating this procedure for vectors C and D  we obtain the following, where × signifies the multiplicative operator and not the cross product: Vector B = ||B|| ×(<0,70,115>)/(25√(29)), Vector C= ||C|| × (<-60,0,115>/(5√673), and Vector=||D||×(<45, -65, 115>)/(5√779).

Now in order for the system to be in Mechanical Equilibrium the net sum of Force vectors must be equivalent to 0 which means B+C+D+W= 0, or B+C+D=-W. With this parameter in mind we can map the vectors into an augmented matrix and solve for each magnitude of each cable to obtain the amount of tension in each cable, which of course depends on the length of the cable in this context. The matrix looks will look like this:

After using a calculator to go through the steps of Gaussian Elimination and putting the matrix into Row Reduced Echelon Form, we find that the tension in AB is ≈ 229.704 Newtons, AC≈178.753 Newtons, and the tension in AD ≈256.421 Newtons