General Chemistry (4th Edition)

Published by University Science Books
ISBN 10: 1891389602
ISBN 13: 978-1-89138-960-3

Chapter 14 Thermochemistry - Problems - Page 509: 3

Answer

$- {\text{8}}{\text{.2}} \times {\text{1}}{{\text{0}}^4}{\text{ J}}$

Work Step by Step

\[\begin{gathered} {\mathbf{Known}} \hfill \\ \hfill \\ {\text{Volume of gas produce = 0}}{\text{.75 }}{{\text{m}}^3} \hfill \\ {\text{Pressure (constant) = 110}}{\text{.0 kPa}} \hfill \\ {\text{Work done by the gas (J) = ?}} \hfill \\ \hfill \\ {\mathbf{Calculations}} \hfill \\ \hfill \\ {\text{Using the formula }} \hfill \\ {\text{w = }} - {\text{p}}\Delta {\text{V}} \hfill \\ {\text{ = }} - {\text{(110}}{\text{.0kPa)(0}}{\text{.75 }}{{\text{m}}^3} - {\text{ 0 }}{{\text{m}}^3}) \hfill \\ {\text{ = }} - {\text{82}}{\text{.5 kPa}} \cdot {{\text{m}}^3} \hfill \\ {\text{The initial volume is 0 }}{{\text{m}}^3}{\text{ because initially,}} \hfill \\ {\text{there is no gaseous product}}{\text{.}} \hfill \\ \hfill \\ {\text{Since 1 Joule is equal to 1 Pa - }}{{\text{m}}^3},{\text{ so}} \hfill \\ {\text{w = }} - {\text{82}}{\text{.5 kPa}} \cdot {{\text{m}}^3} \times \dfrac{{1000{\text{ Pa}}}}{{{\text{1kPa}}}} \hfill \\ {\text{w = }} - 82500{\text{ J }} \approx {\text{ }}\boxed{ - {\text{8}}{\text{.2}} \times {\text{1}}{{\text{0}}^4}{\text{ J}}} \hfill \\ \end{gathered} \]
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