https://ejercicios-fyq.com/ Ejercicios Resueltos, Situaciones de aprendizaje y VÍDEOS de Física y Química para Secundaria y Bachillerato es SPIP - www.spip.net https://www.ejercicios-fyq.com/local/cache-vignettes/L144xH25/siteon0-da713.png?1758361862 https://ejercicios-fyq.com/ 25 144 https://www.ejercicios-fyq.com/General-gas-law-8316 https://www.ejercicios-fyq.com/General-gas-law-8316 2024-09-17T02:52:07Z text/html es F_y_Q State equation Gas laws SOLVED <p>Calculate the final temperature of a gas enclosed in a volume of 2 L at 1 atm, if we reduce its volume to 0.5 L and its pressure increases to 3.8 atm.</p> - <a href="https://www.ejercicios-fyq.com/Matter-and-gas-laws" rel="directory">Matter and gas laws</a> / <a href="https://www.ejercicios-fyq.com/State-equation" rel="tag">State equation</a>, <a href="https://www.ejercicios-fyq.com/Gas-laws" rel="tag">Gas laws</a>, <a href="https://www.ejercicios-fyq.com/SOLVED" rel="tag">SOLVED</a> <div class='rss_texte'><p>Calculate the final temperature of a gas enclosed in a volume of 2 L at 1 atm, if we reduce its volume to 0.5 L and its pressure increases to 3.8 atm.</p></div> <hr /> <div <div class='rss_ps'><p>You will use the state equation of gases: <br/> <br/> <img src='https://www.ejercicios-fyq.com/local/cache-TeX/ffed963df0c5c03aa73305d0c53d9676.png' style="vertical-align:middle;" width="171" height="51" alt="\color[RGB]{2,112,20}{\bm{\frac{P_1\cdot V_1}{T_1} = \frac{P_2\cdot V_2}{T_2}}}}" title="\color[RGB]{2,112,20}{\bm{\frac{P_1\cdot V_1}{T_1} = \frac{P_2\cdot V_2}{T_2}}}}" /> <br/> <br/> Solving for the value of <img src='https://www.ejercicios-fyq.com/local/cache-TeX/6a058d102910f33a7d4cf9ea23067b8c.png' style="vertical-align:middle;" width="23" height="40" alt="T_2" title="T_2" />: <br/> <br/> <p class="spip" style="text-align: center;"><img src='https://www.ejercicios-fyq.com/local/cache-TeX/fc4c300791b1c9df86ae6ffb2261b125.png' style="vertical-align:middle;" width="514" height="50" alt="T_2 = \frac{P_2\cdot V_2\cdot T_1}{P_1\cdot V_1} = \frac{3.8\ \cancel{atm}\cdot 0.5\ \cancel{L}\cdot 298\ K}{1\ \cancel{atm}\cdot 2\ \cancel{L}} = \fbox{\color[RGB]{192,0,0}{\bf 283.1\ K}}" title="T_2 = \frac{P_2\cdot V_2\cdot T_1}{P_1\cdot V_1} = \frac{3.8\ \cancel{atm}\cdot 0.5\ \cancel{L}\cdot 298\ K}{1\ \cancel{atm}\cdot 2\ \cancel{L}} = \fbox{\color[RGB]{192,0,0}{\bf 283.1\ K}}" /></p> </math></p></div> https://www.ejercicios-fyq.com/Application-of-Graham-s-law-hydrogen-diffusion-rate-8308 https://www.ejercicios-fyq.com/Application-of-Graham-s-law-hydrogen-diffusion-rate-8308 2024-09-12T04:06:10Z text/html es F_y_Q Gas laws Graham's law SOLVED <p>Determine the diffusion rate of hydrogen, knowing that the diffusion rate of oxygen is 2 minutes.</p> - <a href="https://www.ejercicios-fyq.com/Matter-and-gas-laws" rel="directory">Matter and gas laws</a> / <a href="https://www.ejercicios-fyq.com/Gas-laws" rel="tag">Gas laws</a>, <a href="https://www.ejercicios-fyq.com/Graham-s-law" rel="tag">Graham's law</a>, <a href="https://www.ejercicios-fyq.com/SOLVED" rel="tag">SOLVED</a> <div class='rss_texte'><p>Determine the diffusion rate of hydrogen, knowing that the diffusion rate of oxygen is 2 minutes.</p></div> <hr /> <div <div class='rss_ps'><p>Graham's law relates the diffusion rates of two gases to their molecular masses. If the gases are A and B, the relationship is: <br/> <br/> <img src='https://www.ejercicios-fyq.com/local/cache-TeX/4cf3fe04d22fc72e2d726e95ac2bdc33.png' style="vertical-align:middle;" width="126" height="65" alt="\color[RGB]{2,112,20}{\bm{\frac{v_A}{v_B} = \sqrt{\frac{M_B}{M_A}}}}" title="\color[RGB]{2,112,20}{\bm{\frac{v_A}{v_B} = \sqrt{\frac{M_B}{M_A}}}}" /> <br/> <br/> For hydrogen and oxygen, both being diatomic, the molecular masses are: <br/> <br/> <img src='https://www.ejercicios-fyq.com/local/cache-TeX/fa228ad48fcef891303029aae15bb4aa.png' style="vertical-align:middle;" width="179" height="52" alt="\left \ce{H2}: 2\cdot 1 = {\color[RGB]{0,112,192}{\bf 2\ u}} \atop \ce{O2}: 2\cdot 16 = {\color[RGB]{0,112,192}{\bf 32\ u}} \right \}" title="\left \ce{H2}: 2\cdot 1 = {\color[RGB]{0,112,192}{\bf 2\ u}} \atop \ce{O2}: 2\cdot 16 = {\color[RGB]{0,112,192}{\bf 32\ u}} \right \}" /> <br/> <br/> The relationship between their diffusion rates will be: <br/> <br/> <img src='https://www.ejercicios-fyq.com/local/cache-TeX/d7383bc419efafca3c82287879aa12d2.png' style="vertical-align:middle;" width="236" height="55" alt="\frac{v_{\ce{H_2}}}{v_{\ce{O_2}}}= \sqrt{\frac{32\ \cancel{u}}{2\ \cancel{u}}} = \sqrt{16} = \color[RGB]{0,112,192}{\bf 4}" title="\frac{v_{\ce{H_2}}}{v_{\ce{O_2}}}= \sqrt{\frac{32\ \cancel{u}}{2\ \cancel{u}}} = \sqrt{16} = \color[RGB]{0,112,192}{\bf 4}" /> <br/> <br/> This means hydrogen diffuses four times faster than oxygen, so <b>the diffusion rate of hydrogen will be 0.5 minutes</b>, or 30 seconds</b>.</math></p></div> https://www.ejercicios-fyq.com/General-gas-law-final-volume-of-a-gas-1927 https://www.ejercicios-fyq.com/General-gas-law-final-volume-of-a-gas-1927 2012-11-05T07:22:34Z text/html es F_y_Q EDICO State equation General gas law Gas laws SOLVED <p>Calculate the final volume of a gas at and 1.1 atm, when its volume is 3 L and the temperature and pressure are increased to and 1.3 atm.</p> - <a href="https://www.ejercicios-fyq.com/Matter-and-gas-laws" rel="directory">Matter and gas laws</a> / <a href="https://www.ejercicios-fyq.com/EDICO" rel="tag">EDICO</a>, <a href="https://www.ejercicios-fyq.com/State-equation" rel="tag">State equation</a>, <a href="https://www.ejercicios-fyq.com/General-gas-law" rel="tag">General gas law</a>, <a href="https://www.ejercicios-fyq.com/Gas-laws" rel="tag">Gas laws</a>, <a href="https://www.ejercicios-fyq.com/SOLVED" rel="tag">SOLVED</a> <div class='rss_texte'><p>Calculate the final volume of a gas at <img src='https://www.ejercicios-fyq.com/local/cache-vignettes/L55xH42/e377e6d6fa467c168c039e4ec52eff00-8b3d5.png?1732957639' style='vertical-align:middle;' width='55' height='42' alt="30\ ^oC" title="30\ ^oC" /> and 1.1 atm, when its volume is 3 L and the temperature and pressure are increased to <img src='https://www.ejercicios-fyq.com/local/cache-vignettes/L52xH17/63a42815bec477dc118cf61ee35689fc-6eeae.png?1732957639' style='vertical-align:middle;' width='52' height='17' alt="42\ ^oC" title="42\ ^oC" /> and 1.3 atm.</math></p></div> <hr /> <div <div class='rss_ps'><p>First, isolate the final volume in the general gas equation: <br/> <br/> <img src='https://www.ejercicios-fyq.com/local/cache-TeX/0e536212229ed24e2f72242063e33f3f.png' style="vertical-align:middle;" width="366" height="51" alt="\frac{P_1\cdot V_1}{T_1} = \frac{P_2\cdot V_2}{T_2}\ \to\ \color[RGB]{2,112,20}{\bm{V_2= \frac{P_1\cdot V_1\cdot T_2}{P_2\cdot T_1}}}" title="\frac{P_1\cdot V_1}{T_1} = \frac{P_2\cdot V_2}{T_2}\ \to\ \color[RGB]{2,112,20}{\bm{V_2= \frac{P_1\cdot V_1\cdot T_2}{P_2\cdot T_1}}}" /> <br/> <br/> Replace the given data in the equation. Remember that the temperature must be expressed on the Kelvin scale: <p class="spip" style="text-align: center;"><img src='https://www.ejercicios-fyq.com/local/cache-TeX/6a568b6f1f7c57047becd5d7b56e31af.png' style="vertical-align:middle;" width="352" height="48" alt="V_2= \frac{1.1\ \cancel{atm}\cdot 3\ L\cdot 315\ \cancel{K}}{1.3\ \cancel{atm}\cdot 303\ \cancel{K}} = \fbox{\color[RGB]{192,0,0}{\bf 2.64\ L}}" title="V_2= \frac{1.1\ \cancel{atm}\cdot 3\ L\cdot 315\ \cancel{K}}{1.3\ \cancel{atm}\cdot 303\ \cancel{K}} = \fbox{\color[RGB]{192,0,0}{\bf 2.64\ L}}" /></p> </math></p> <p> <br/></p> <p><b><a href="https://ejercicios-fyq.com/Situaciones-de-aprendizaje/EDICO/Ej_1927.edi" download>Download the problem statement and solution in EDICO format if needed.</a></b></p></div>