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Chapter 6: Problem 28

A 12.8 L cylinder contains \(35.8 \mathrm{g} \mathrm{O}_{2}\) at \(46^{\circ} \mathrm{C}\). What is the pressure of this gas, in atmospheres?

Short Answer

Expert verified
The pressure of the gas is \(2.29 \, \mathrm{atm}\).

Step by step solution

01

Convert grams of \(O_2\) to moles

We first need to convert the mass of \(O_2\) (oxygen) in grams into moles, by using the molar mass of \(O_2\), which is \(32.00 \, \mathrm{g/mol}\). This is done using the formula \(n = m/M\), where \(m\) is the mass and \(M\) is the molar mass. This gives us: \[n = 35.8 \, \mathrm{g} / 32.00 \, \mathrm{g/mol} = 1.119 \, \mathrm{mol}\]
02

Convert temperature from Celsius to Kelvin

We need to convert the temperature from Celsius to Kelvin, because the ideal gas law uses the Kelvin temperature scale. We can do this using the formula \(T(K) = T(C) + 273.15\), which gives us: \[T = 46^{\circ}C + 273.15 = 319.15 \, K\]
03

Substitute values into the ideal gas law and solve for P

Next, we put these values into the ideal gas law equation \(PV = nRT\), and solve for the pressure \(P\). Remembering that the ideal gas constant \(R = 0.0821 \, \mathrm{(atm \cdot L)/(mol \cdot K)}\), we get: \[P = \frac{nRT}{V} = \frac{1.119 \, \mathrm{mol} \cdot 0.0821 \, \mathrm{(atm \cdot L)/(mol \cdot K)} \cdot 319.15 \, K}{12.8 \, L} = 2.29 \, \mathrm{atm}\] This pressure is the final answer to the exercise.

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Most popular questions from this chapter

If the Kelvin temperature of a sample of ideal gas doubles (e.g., from 200 K to 400 K), what happens to the root-mean-square speed, \(u_{\mathrm{rms}}\) ? (a) \(u_{\mathrm{rms}}\) increases by a factor of \(\sqrt{2} ;\) (b) \(u_{\mathrm{rms}}\) increases by a factor of \(2 ;(\mathrm{c}) u_{\mathrm{rms}}\) decreases by a factor of 2 (d) \(u_{\mathrm{rms}}\) increases by a factor of \(4 ;\) (e) \(u_{\mathrm{rms}}\) decreases by a factor of 4.

In the reaction of \(\mathrm{CO}_{2}(\mathrm{g})\) and solid sodium peroxide \(\left(\mathrm{Na}_{2} \mathrm{O}_{2}\right),\) solid sodium carbonate \(\left(\mathrm{Na}_{2} \mathrm{CO}_{3}\right)\) and oxy- gen gas are formed. This reaction is used in submarines and space vehicles to remove expired \(\mathrm{CO}_{2}(\mathrm{g})\) and to generate some of the \(\mathrm{O}_{2}(\mathrm{g})\) required for breathing. Assume that the volume of gases exchanged in the lungs equals \(4.0 \mathrm{L} / \mathrm{min},\) the \(\mathrm{CO}_{2}\) content of expired air is \(3.8 \% \mathrm{CO}_{2}\) by volume, and the gases are at \(25^{\circ} \mathrm{C}\) and \(735 \mathrm{mmHg}\). If the \(\mathrm{CO}_{2}(\mathrm{g})\) and \(\mathrm{O}_{2}(\mathrm{g})\) in the above reaction are measured at the same temperature and pressure, (a) how many milliliters of \(\mathrm{O}_{2}(\mathrm{g})\) are produced per minute and \((\mathrm{b})\) at what rate is the \(\mathrm{Na}_{2} \mathrm{O}_{2}(\mathrm{s})\) consumed, in grams per hour?

Explain why it is necessary to include the density of \(\mathrm{Hg}(1)\) and the value of the acceleration due to gravity, \(g,\) in a precise definition of a millimeter of mercury (page 194 ).

A mixture of \(\mathrm{H}_{2}(\mathrm{g})\) and \(\mathrm{O}_{2}(\mathrm{g})\) is prepared by electrolyzing \(1.32 \mathrm{g}\) water, and the mixture of gases is collected over water at \(30^{\circ} \mathrm{C}\) and \(748 \mathrm{mmHg} .\) The volume of "wet" gas obtained is 2.90 L. What must be the vapor pressure of water at \(30^{\circ} \mathrm{C} ?\) $$2 \mathrm{H}_{2} \mathrm{O}(1) \stackrel{\text { electrolysis }}{\longrightarrow} 2 \mathrm{H}_{2}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{g})$$

A compound is \(85.6 \%\) carbon by mass. The rest is hydrogen. When \(10.0 \mathrm{g}\) of the compound is evaporated at \(50.0^{\circ} \mathrm{C},\) the vapor occupies \(6.30 \mathrm{L}\) at \(1.00 \mathrm{atm}\) pressure. What is the molecular formula of the compound?
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