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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 temp

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1056
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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?

May 31, 2015

#2
+1036
+10

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

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

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

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Jun 1, 2015

#1
+27374
+5

See Nauseated's solution below.

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Jun 1, 2015
#2
+1036
+10

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

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

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

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Nauseated Jun 1, 2015
#3
+95356
0

Thanks Nauseated,

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

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

Jun 1, 2015