Nauseated

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$$\text {}\\
\noindent \text {Set up simultaneous equations using known data set and relations:} \\\\
\noindent \text {P*V = nRT }\\
\text {P = pressure in Pa}\\
\text {V = volume in } m^3\\
\text {T = in Kelvin} \\
\text {n = moles} \\
\text {R = 8.314J mol}^{-1}\\\\
\noindent P2 =P1 + 800 \\
T2 = T1 - 8 \\\\
\noindent P1*(2.5E-2) = 4.1 * 8.314 * T1 \\
\noindent (P1 + 800)*(1.9E-2) = 4.1 * 8.314 *(T1-8) \\\\
\text {Using the site calculator to solve:} \\\\$$

$$\underset{\,\,\,\,{\textcolor[rgb]{0.66,0.66,0.66}{\rightarrow {\mathtt{p1, t1}}}}}{{solve}}{\left(\begin{array}{l}{\mathtt{P1}}{\mathtt{\,\times\,}}{\mathtt{0.025}}={\mathtt{4.1}}{\mathtt{\,\times\,}}{\mathtt{8.314}}{\mathtt{\,\times\,}}{\mathtt{T1}}\\
\left({\mathtt{P1}}{\mathtt{\,\small\textbf+\,}}{\mathtt{800}}\right){\mathtt{\,\times\,}}{\mathtt{0.019}}={\mathtt{4.1}}{\mathtt{\,\times\,}}{\mathtt{8.314}}{\mathtt{\,\times\,}}\left({\mathtt{T1}}{\mathtt{\,-\,}}{\mathtt{8}}\right)\end{array}\right)} \Rightarrow \left\{ \begin{array}{l}{\mathtt{p1}} = {\frac{{\mathtt{239\,916}}}{{\mathtt{5}}}}\\
{\mathtt{t1}} = {\frac{{\mathtt{5\,997\,900}}}{{\mathtt{170\,437}}}}\\
\end{array} \right\} \Rightarrow \left\{ \begin{array}{l}{\mathtt{p1}} = {\mathtt{47\,983.2}}\\
{\mathtt{t1}} = {\mathtt{35.191\: \!302\: \!358\: \!056\: \!055\: \!9}}\\
\end{array} \right\}$$

$$\text {}\\
\noindent \text {The original pressure of the gas } 47983 { Pa} \\
\text {The original temperature of the gas } 35.91 { K}\\$$

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Classify it under Rube Goldberg Solutions for complex math problems with trivial non-roman ZEROS.

Sub-classification: Put a Sock in it.

Rosala, it’s rare anyone likes and appreciates a troll. It’s more rare that someone will publicly acknowledge this with a genuine and refined compliment.

Thank you.

Melody taught me about Permutations, too. She's an expert at per-mutating very funny troll-posts into Scooby Snacks.

She has a special jar where she keeps them.

Nauseated......your monkeys weren't "monkeying around" on here to cause that crash, were they ???

I wondered if that might be the case, CPhill, so I sent an email to Lancelot Link asking if he knew anything about 2.0calc’s logon and post failure. To my surprise Chimp Loki replied, instead of Lancelot.

He said, "Careful investigation of the evidence shows CPhill posted on both occasions just before failure occurred. While it’s not conclusive, it indicates a major contribution to the system’s malfunction. . . .."  

He added: “. . . one doesn’t need to be a monkey to monkey around.”

Looks very incriminating to me . . . What do you think, Melody?

Now I can get back to trolling. . . . I mean answering math questions in a proper manner.

Hot D**n! The ghosts are exorcised  . . .

Let’s see . . . Toxic fumes from burning Titanium . . . just a guess.

  Nero was fiddling . . . so it’s a good guess.

Zac's hemisphere and my hemisphere had a bottom on it.

Yes, I noticed. The bottom is part of the lower half of a CDD Sphere.

A hemisphere is one half of a sphere. Doubling this a sphere. If the bottom area of a hemisphere is needed then it’s calculated separately. In this case it’s 308.965cm²

Calculating the radius from a hemisphere that includes the area of the bottom returns a value that will correlate to a surface area value that is smaller than the sum of two identical hemispheres.

Our brains have ceribral hemispheres.

By way of curiosity, which half were you using?

Mine is full of grey matter. Does yours is contains nothing?

I am unsure of the color, but the callosal commissure joins two cerebral CDD hemispheres. However, in this case, the whole is greater than the sum of its parts –instead of less than. :)

$$\displaystyle \\ \text {Surface area of a hemisphere is }\\

\noindent A_ s = 2\pi*r^2\\\

\noindent \text {Radius is }\\

r = \sqrt{\dfrac{A_s}{ 2\pi}}\hspace{30pt}| A_s=618cm^2 \\\

r = \sqrt{\dfrac{618cm^2}{ 2\pi}}\; =\; 9.917cm \\\\

\text {The other answers are \rm \pm $ the upper and lower half of a CDD Sphere.}$$