Chapter 10: Problem 1
Mars has an average atmospheric pressure of 0.007 atm. Would it be easier or harder to drink from a straw on Mars than on Earth? Explain. [Section 10.2\(]\)
Chapter 10: Problem 1
Mars has an average atmospheric pressure of 0.007 atm. Would it be easier or harder to drink from a straw on Mars than on Earth? Explain. [Section 10.2\(]\)
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Get started for freeNewton had an incorrect theory of gases in which he assumed that all gas molecules repel one another and the walls of their container. Thus, the molecules of a gas are statically and uniformly distributed, trying to get as far apart as possible from one another and the vessel walls. This repulsion gives rise to pressure. Explain why Charles's law argues for the kineticmolecular theory and against Newton's model.
You have a gas confined to a cylinder with a movable piston. What would happen to the gas pressure inside the cylinder if you do the following? (a) Decrease the volume to one-fourth the original volume while holding the temperature constant. (b) Reduce the temperature (in kelvins) to half its original value while holding the volume constant. (c) Reduce the amount of gas to one-fourth while keeping the volume and temperature constant.
(a) Place the following gases in order of increasing average molecular speed at \(300 \mathrm{~K}: \mathrm{CO}, \mathrm{SF}_{6}, \mathrm{H}_{2} \mathrm{~S}, \mathrm{Cl}_{2}, \mathrm{HBr}\). (b) Calcu- late and compare the rms speeds of \(\mathrm{CO}\) and \(\mathrm{Cl}_{2}\) molecules at \(300 \mathrm{~K} .(\mathbf{c})\) Calculate and compare the most probable speeds of \(\mathrm{CO}\) and \(\mathrm{Cl}_{2}\) molecules at \(300 \mathrm{~K}\).
The metabolic oxidation of glucose, \(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6},\) in our bodies produces \(\mathrm{CO}_{2},\) which is expelled from our lungs as a gas: $$ \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}(a q)+6 \mathrm{O}_{2}(g) \longrightarrow 6 \mathrm{CO}_{2}(g)+6 \mathrm{H}_{2} \mathrm{O}(l) $$ (a) Calculate the volume of dry \(\mathrm{CO}_{2}\) produced at body temperature \(\left(37^{\circ} \mathrm{C}\right)\) and 0.970 atm when \(24.5 \mathrm{~g}\) of glucose is consumed in this reaction. (b) Calculate the volume of oxygen you would need, at 1.00 atm and \(298 \mathrm{~K}\), to completely oxidize \(50.0 \mathrm{~g}\) of glucose.
The physical fitness of athletes is measured by " \(V_{\mathrm{O}_{2}}\) max," which is the maximum volume of oxygen consumed by an individual during incremental exercise (for example, on a treadmill). An average male has a \(V_{\mathrm{O}_{2}}\) max of \(45 \mathrm{~mL} \mathrm{O}_{2} / \mathrm{kg}\) body mass \(/ \mathrm{min}\), but a world-class male athlete can have a \(V_{\mathrm{O}_{2}}\) max reading of \(88.0 \mathrm{~mL} \mathrm{O}_{2} / \mathrm{kg}\) body mass \(/ \mathrm{min.}\) (a) Calculate the volume of oxygen, in mL, consumed in 1 hr by an average man who weighs 185 lbs and has a \(V_{\mathrm{O}_{2}}\) max reading of 47.5 \(\mathrm{mL} \mathrm{O}_{2} / \mathrm{kg}\) body mass \(/ \mathrm{min}\) (b) If this man lost \(20 \mathrm{lb}\), exercised, and increased his \(V_{\mathrm{O}_{2}}\) max to \(65.0 \mathrm{~mL} \mathrm{O}_{2} / \mathrm{kg}\) body mass \(/ \mathrm{min}\), how many mL of oxygen would he consume in \(1 \mathrm{hr}\) ?
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