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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 \(0.966\) atm. (b) The mercury in the arm attached to the gas is \(8.7 \mathrm{~mm}\) lower than in the one open to the atmosphere; atmospheric pressure is \(0.99\) atm.

Short Answer

Expert verified
The pressure of the enclosed gas in torr for case (a) is \(750.36 \ \text{torr}\), and for case (b) is \(743.7 \ \text{torr}\).

Step by step solution

01

Case (a): Mercury height in gas arm is higher

First, convert atmospheric pressure to torr: Atmospheric pressure = \(0.966 \ \text{atm}\) 1 atm = \(760 \ \text{torr}\) Atmospheric pressure = \(0.966 \times 760 \ \text{torr} = 734.96 \ \text{torr}\) Next, calculate pressure due to height difference of mercury: Height difference = \(15.4 \ \text{mm}\) 1 mmHg = 1 torr Pressure difference = \(15.4 \ \text{torr}\) Now, find the gas pressure in the container by adding the atmospheric pressure and the pressure difference: Gas pressure = atmospheric pressure + pressure difference = \(734.96 \ \text{torr} + 15.4 \ \text{torr} = 750.36 \ \text{torr}\)
02

Case (b): Mercury height in gas arm is lower

First, convert atmospheric pressure to torr: Atmospheric pressure = \(0.99 \ \text{atm}\) 1 atm = \(760 \ \text{torr}\) Atmosphere pressure = \(0.99 \times 760 \ \text{torr} = 752.4 \ \text{torr}\) Next, calculate pressure due to the height difference of mercury: Height difference = \(8.7 \ \text{mm}\) 1 mmHg = 1 torr Pressure difference = \(8.7 \ \text{torr}\) Now, find the gas pressure in the container by subtracting the pressure difference from the atmospheric pressure: Gas pressure = atmospheric pressure - pressure difference = \(752.4 \ \text{torr} - 8.7 \ \text{torr} = 743.7 \ \text{torr}\) Therefore, the pressure of the enclosed gas in torr for case (a) is 750.36 torr, and for case (b) is 743.7 torr.

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Key Concepts

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

Manometer
A manometer is a widely used device to measure gas pressure. It consists of a U-shaped tube filled with a liquid, commonly mercury. This tool can measure the difference in pressure between the gas in a container and the atmospheric pressure outside. Manometers can be open-end or closed-end.
Open-end manometers allow the liquid's free arm to be exposed to the atmosphere, whereas closed-end manometers have one end sealed.
When a gas is introduced to the manometer, it displaces the liquid vertically. These changes reflect differences in pressure.
  • If the liquid's level in the tube is higher on the side attached to the gas, the gas pressure is higher than the atmospheric pressure.
  • Conversely, if it's lower, the gas pressure is lower than the atmospheric pressure.
    These variations allow us to calculate the gas's precise pressure by considering the height difference in the liquid column.
Atmospheric Pressure Conversion
Atmospheric pressure is usually measured in atmospheres (atm), but other units like torr and millimeters of mercury (mmHg) are also common.
To solve problems involving manometers, we often need to convert atmospheric pressure to different units to match the units of the pressure difference.
The conversion between atmospheres and torr is straightforward because 1 atm is equal to 760 torr. This relationship is used to convert atmospheric pressure to torr, essential for calculations involving manometers. For example, converting 0.966 atm to torr involves multiplying by 760, resulting in approximately 734.96 torr.
  • Always ensure conversion accuracy to avoid errors in further calculations.
  • This conversion step is crucial before moving on to calculating pressure differences.
Pressure Difference Calculation
Calculating the pressure difference is a key step in determining the gas pressure inside a container using a manometer.
The pressure difference is derived from the height difference of the liquid column.
The relationship here is simple: 1 mmHg corresponds directly to 1 torr. This implies that a 15 mm difference in the mercury column equates to a 15 torr pressure difference.
By using this relation, we can:
  • Add the pressure difference to atmospheric pressure if the liquid level is higher on the side connected to the gas.
  • Subtract the pressure difference from atmospheric pressure if the liquid level is lower on the side attached to the gas.
    This step, along with atmospheric pressure conversion, allows for the precise computation of the gas's pressure.

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Most popular questions from this chapter

In the United States, barometric pressures are generally reported in inches of mercury (in. \(\mathrm{Hg}\) ). On a beautiful summer day in Chicago the barometric pressure is \(30.45\) in. Hg. (a) Convert this pressure to torr. (b) A meteorologist explains the nice weather by referring to a "high-pressure area." In light of your answer to part (a), explain why this term makes sense.

Calculate each of the following quantities for an ideal gas: (a) the volume of the gas, in liters, if \(1.50 \mathrm{~mol}\) has a pressure of \(0.985\) atm at a temperature of \(-6{ }^{\circ} \mathrm{C} ;\) (b) the absolute temperature of the gas at which \(3.33 \times 10^{-3} \mathrm{~mol}\) occupies \(325 \mathrm{~mL}\) at 750 torr; \((\mathrm{c})\) the pressure, in atmospheres, if \(0.0467 \mathrm{~mol}\) occupies \(413 \mathrm{~mL}\) at \(138^{\circ} \mathrm{C} ;\) (d) the quantity of gas, in moles, if \(55.7\) Lat \(54{ }^{\circ} \mathrm{C}\) has a pressure of \(11.25 \mathrm{kPa}\).

Calcium hydride, \(\mathrm{CaH}_{2}\), reacts with water to form hydrogen gas: $$ \mathrm{CaH}_{2}(s)+2 \mathrm{H}_{2} \mathrm{O}(I) \longrightarrow \mathrm{Ca}(\mathrm{OH})_{2}(a q)+2 \mathrm{H}_{2}(g) $$ This reaction is sometimes used to inflate life rafts, weather balloons, and the like, where a simple, compact means of generating \(\mathrm{H}_{2}\) is desired. How many grams of \(\mathrm{CaH}_{2}\) are needed to generate \(53.5 \mathrm{~L}\) of \(\mathrm{H}_{2}\) gas if the pressure of \(\mathrm{H}_{2}\) is 814 torr at \(21{ }^{\circ} \mathrm{C}\) ?

Briefly explain the significance of the constants \(a\) and \(b\) in the van der Waals equation.

(a) How is the law of combining volumes explained by Avogadro's hypothesis? (b) Consider a 1.0-L flask containing neon gas and a 1.5-L flask containing xenon gas. Both gases are at the same pressure and temperature. According to Avogadro's law, what can be said about the ratio of the number of atoms in the two flasks?

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