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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?

difficulty advanced
Guest May 31, 2015

Best Answer 

 #2
avatar+1035 
+10

.

 

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

 

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

 

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

 

.

Nauseated  Jun 1, 2015
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3+0 Answers

 #1
avatar+26329 
+5

See Nauseated's solution below.

.

Alan  Jun 1, 2015
 #2
avatar+1035 
+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
 #3
avatar+91049 
0

Thanks Nauseated,

Nice use of the site calc to solve the equations simultaneously too.   

I always forget how to do things like that. :)

Melody  Jun 1, 2015

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