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An open-end manometer containing mercury is connected to a container of gas, as depicted in Sample Exercise \(10.2 .\) What is the pressure of the enclosed gas in torr in each of the following situations? (a) The mercury in the arm attached to the gas is \(15.4 \mathrm{~mm}\) higher than in the one open to the atmosphere; atmospheric pressure is 0.985 atm. (b) The mercury in the arm attached to the gas is \(12.3 \mathrm{~mm}\) lower than in the one open to the atmosphere; atmospheric pressure is \(0.99 \mathrm{~atm} .\)

Short Answer

Expert verified
In Situation (a), the pressure of the enclosed gas is 1017.45 torr. In Situation (b), the pressure of the enclosed gas is 536.63 torr.

Step by step solution

01

(Situation a: Convert atmospheric pressure to torr)

Firstly, we need to convert the atmospheric pressure given in atm to units of torr. 1 atm is equal to 760 torr, so let's convert 0.985 atm to torr: \[0.985 \ \mathrm{atm} \times \frac{760 \ \mathrm{torr}}{1 \ \mathrm{atm}} = 748.1 \ \mathrm{torr}\]
02

(Situation a: Calculate height difference in pressure)

Now, we need to convert the height difference in mercury levels to pressure difference. Given that the density of mercury (\(蟻\)) is 13.6 g/cm鲁 and gravitational acceleration (\(g\)) is 980 cm/s虏, the pressure difference (\(螖P\)) can be calculated using the formula \(螖P = 蟻gh\), where h is the height difference in centimeters (converted from mm). Here, h = 1.54 cm (as 15.4 mm is equal to 1.54 cm): \[螖P = 13.6 \frac{\mathrm{g}}{\mathrm{cm}^3} \times 980 \frac{\mathrm{cm}}{\mathrm{s}^2} \times 1.54 \ \mathrm{cm} = 20470.56 \frac{\mathrm{dyn}}{\mathrm{cm}^2}\] To convert dynes per square centimeter to torr, we will divide by 76.0 dynes per square centimeter (as 1 torr = 76.0 dynes per square centimeter): \[\frac{20470.56 \frac{\mathrm{dyn}}{\mathrm{cm}^2}}{76.0 \ \mathrm{dyn} \ \mathrm{cm}^{-2} \ \mathrm{torr}^{-1}} = 269.35 \ \mathrm{torr}\]
03

(Situation a: Calculate gas pressure)

Finally, we need to calculate the pressure of the gas in the container. Since the mercury level in the arm attached to the gas is higher than the one open to the atmosphere, the gas pressure would be higher than the atmospheric pressure. We will add the pressure difference we calculated to the atmospheric pressure to get the enclosed gas pressure: \[P_\mathrm{gas} = P_\mathrm{atm} + 螖P = 748.1 \ \mathrm{torr} + 269.35 \ \mathrm{torr} = 1017.45 \ \mathrm{torr}\] In Situation (a), the pressure of the enclosed gas is 1017.45 torr.
04

(Situation b: Convert atmospheric pressure to torr)

Just like in Situation (a), we need to convert the atmospheric pressure given in atm to units of torr. To do that, we multiply 0.99 atm with 760 torr/atm: \[0.99 \ \mathrm{atm} \times \frac{760 \ \mathrm{torr}}{1 \ \mathrm{atm}} = 752.4 \ \mathrm{torr}\]
05

(Situation b: Calculate height difference in pressure)

Now, we need to convert the height difference in mercury levels to pressure difference. Here, h = -1.23 cm (as 12.3 mm is equal to 1.23 cm and we use a negative sign since the mercury in the arm attached to the gas is lower than the one open to the atmosphere): \[螖P = 13.6 \frac{\mathrm{g}}{\mathrm{cm}^3} \times 980 \frac{\mathrm{cm}}{\mathrm{s}^2} \times (-1.23) \ \mathrm{cm} = -16398.72 \frac{\mathrm{dyn}}{\mathrm{cm}^2}\] To convert dynes per square centimeter to torr, we will divide by 76.0 dynes per square centimeter: \[\frac{-16398.72 \frac{\mathrm{dyn}}{\mathrm{cm}^2}}{76.0 \ \mathrm{dyn} \ \mathrm{cm}^{-2} \ \mathrm{torr}^{-1}} = -215.77 \ \mathrm{torr}\]
06

(Situation b: Calculate gas pressure)

Finally, we need to calculate the pressure of the gas in the container. Since the mercury level in the arm attached to the gas is lower than the one open to the atmosphere, the gas pressure would be lower than the atmospheric pressure. We will add the pressure difference we calculated to the atmospheric pressure to get the enclosed gas pressure: \[P_\mathrm{gas} = P_\mathrm{atm} + 螖P = 752.4 \ \mathrm{torr} - 215.77 \ \mathrm{torr} = 536.63 \ \mathrm{torr}\] In Situation (b), the pressure of the enclosed gas is 536.63 torr.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Pressure Unit Conversion
When working with gases and pressures, it's common to encounter different units of measurement. One essential skill is converting between these units. Atmospheric pressure, for example, is often measured in atmospheres (atm), but may also need to be represented in torr or millimeters of mercury (mmHg), as these are more suitable for lab measurements.

The conversion factor to remember is that 1 atm is equivalent to 760 torr. Using this relationship allows for the direct conversion of atmospheric pressure values from atm to torr. For instance, an atmospheric pressure of 0.985 atm can be converted to torr by multiplying it by 760, resulting in 748.1 torr, as shown in the situation a of our exercise.

Similarly, for the pressure in situation b, converting 0.99 atm to torr using the multiplication by 760 yields 752.4 torr. Without mastering the unit conversion, solving problems involving different pressure units can become exceptionally challenging. Recognizing the need to convert these values is fundamental and often the first step in solving manometer-related problems.
Mercury Manometer Equations
Understanding the equations that govern a mercury manometer is key to determining gas pressure. The core equation used involves the density of mercury \(\rho = 13.6 \frac{\mathrm{g}}{\mathrm{cm}^3}\), the gravitational acceleration \(g = 980 \frac{\mathrm{cm}}{\mathrm{s}^2}\), and the height difference in the mercury column \(h\). This equation allows us to calculate the pressure difference \(\Delta P\) induced by the height difference using \(\Delta P = \rho gh\).

Once the pressure difference is calculated in dynes per square centimeter, converting it to torr by dividing by the constant 76.0 is necessary since 1 torr equals 76.0 dynes per square centimeter. This conversion is critical, as seen in step 2 of the exercise for situation a, where the calculated difference was 269.35 torr. Understanding these conversions and the relationship between height difference and pressure change enables solving for the gas pressure in the container attached to the manometer.
Atmospheric Pressure Adjustment
In manometry, atmospheric pressure provides a reference point against which the gas pressure is measured. However, the current level of atmospheric pressure must be adjusted in calculations to accurately determine the pressure within a container.

When the mercury column is higher on the side of the gas container, it indicates that the gas pressure exceeds atmospheric pressure. We must add the calculated pressure difference due to the height of the mercury to the atmospheric pressure, as shown in step 3. Conversely, if the mercury column is lower on the gas side than the atmospheric side, it indicates that the gas pressure is less than atmospheric pressure. As seen in step 6 for situation b, we subtract the pressure difference from the atmospheric pressure.

It is critical to adjust for atmospheric pressure correctly to ensure that the final pressure reading of the gas is accurate. For students, recognizing this adjustment provides deeper insight into how a manometer functions and allows more accurate predictions of gas behavior under different pressure conditions.

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Most popular questions from this chapter

What property or properties of gases can you point to that support the assumption that most of the volume in a gas is empty space?

An herbicide is found to contain only \(\mathrm{C}, \mathrm{H}, \mathrm{N},\) and \(\mathrm{Cl} .\) The complete combustion of a 100.0 -mg sample of the herbicide in excess oxygen produces \(83.16 \mathrm{~mL}\) of \(\mathrm{CO}_{2}\) and \(73.30 \mathrm{~mL}\) of \(\mathrm{H}_{2} \mathrm{O}\) vapor at STP. A separate analysis shows that the sample also contains \(16.44 \mathrm{mg}\) of Cl. (a) Determine the percent composition of the substance. (b) Calculate its empirical formula. (c) What other information would you need to know about this compound to calculate its true molecular formula?

A gas of unknown molecular mass was allowed to effuse through a small opening under constant-pressure conditions. It required 105 s for \(1.0 \mathrm{~L}\) of the gas to effuse. Under identical experimental conditions it required \(31 \mathrm{~s}\) for \(1.0 \mathrm{~L}\) of \(\mathrm{O}_{2}\) gas to effuse. Calculate the molar mass of the unknown gas. (Remember that the faster the rate of effusion, the shorter the time required for effusion of \(1.0 \mathrm{~L} ;\) that is, rate and time are inversely proportional.)

Magnesium can be used as a "getter" in evacuated enclosures to react with the last traces of oxygen. (The magnesium is usually heated by passing an electric current through a wire or ribbon of the metal.) If an enclosure of \(0.452 \mathrm{~L}\) has a partial pressure of \(\mathrm{O}_{2}\) of \(3.5 \times 10^{-6}\) torr at \(27{ }^{\circ} \mathrm{C}\), what mass of magnesium will react according to the following equation? $$ 2 \mathrm{Mg}(s)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{MgO}(s) $$

In an experiment reported in the scientific literature, male cockroaches were made to run at different speeds on a miniature treadmill while their oxygen consumption was measured. In one hour the average cockroach running at \(0.08 \mathrm{~km} / \mathrm{hr}\) consumed \(0.8 \mathrm{~mL}\) of \(\mathrm{O}_{2}\) at 1 atm pressure and \(24{ }^{\circ} \mathrm{C}\) per gram of insect mass. (a) How many moles of \(\mathrm{O}_{2}\) would be consumed in 1 hr by a 5.2-g cockroach moving at this speed? (b) This same cockroach is caught by a child and placed in a 1 - qt fruit jar with a tight lid. Assuming the same level of continuous activity as in the research, will the cockroach consume more than \(20 \%\) of the available \(\mathrm{O}_{2}\) in a 48 -hr period? (Air is \(21 \mathrm{~mol}\) percent \(\left.\mathrm{O}_{2} .\right).\)

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