A sealed container containing 4.1 mol of gas is squeezed, changing its volume from 2.5×10−2 m3 to 1.9×10−2 m3. During this process, the temperature decreases by 8.0 K while the pressure increases by 800 Pa.What was the original pressure of the gas in the container?

Guest May 31, 2015

#2**+10 **

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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}\\$$

.

Nauseated
Jun 1, 2015

#2**+10 **

Best Answer

.

$$\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}\\$$

.

Nauseated
Jun 1, 2015