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Chapter 11: Problem 1

The pressure of a gas is \(440 \mathrm{mm}\) Hg. Express this pressure in units of (a) atmospheres, (b) bars, and (c) kilopascals.

Short Answer

Expert verified
0.5789 atm, 0.5866 bar, 58.6621 kPa

Step by step solution

01

- Convert mm Hg to Atmospheres

To convert from mm Hg to atmospheres, use the conversion factor: 1 atm = 760 mm Hg. \[ \text{Pressure in atm} = \frac{440}{760} = 0.5789 \space \text{atm} \]
02

- Convert mm Hg to Bars

To convert from mm Hg to bars, use the conversion factor: 1 mm Hg = 0.00133322 bars. \[ \text{Pressure in bars} = 440 \times 0.00133322 = 0.5866 \space \text{bar} \]
03

- Convert mm Hg to Kilopascals

To convert from mm Hg to kilopascals, use the conversion factor: 1 mm Hg = 0.133322 kPa. \[ \text{Pressure in kPa} = 440 \times 0.133322 = 58.6621 \space \text{kPa} \]

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Key Concepts

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

Atmospheres Conversion
Atmospheric pressure is a key metric for understanding how the weight of air impacts our environment. When you are converting pressure from millimeters of mercury (mm Hg) to atmospheres (atm), you are shifting from a localized measurement to a universal standard. For this process, the conversion factor is critical: 1 atmosphere equals 760 mm Hg.
  • To make the conversion, simply divide the given mm Hg value by 760.
  • For example, if the pressure is 440 mm Hg, you calculate the atmospheres by \( \frac{440}{760} = 0.5789 \space \text{atm} \).
Using atmospheres as a unit can simplify the comparison of pressures and is common in many scientific calculations. Remember, this conversion gives you an understanding of how the pressure compares to the standard atmospheric level at sea level.
Bars Conversion
The bar is another unit of pressure commonly used in meteorology and various applications requiring precise measurement, especially in Europe. To convert from mm Hg to bars, know that 1 mm Hg corresponds to 0.00133322 bars.
  • Start by multiplying the mm Hg value by this bar conversion factor.
  • If you have 440 mm Hg, multiply to find out \( 440 \times 0.00133322 = 0.5866 \space \text{bar} \).
Bars are widely used because the value closely approximates standard atmospheric pressure and is easy to work with, particularly in weather forecasts and for tire pressures.
Kilopascals Conversion
The kilopascal (kPa) is part of the International System of Units (SI) and is frequently used in scientific and engineering contexts. To shift from mm Hg to kPa, use the conversion factor: 1 mm Hg equates to 0.133322 kPa.
  • Calculate the kilopascals by multiplying the mm Hg value by 0.133322.
  • For instance, starting with 440 mm Hg, you'll get \( 440 \times 0.133322 = 58.6621 \space \text{kPa} \).
Kilopascals are particularly convenient for understanding pressures in contexts like gas laws and physical sciences because they directly relate to the metric system, making it easier to integrate into international discussions and applications.

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Most popular questions from this chapter

Ammonia gas is synthesized by combining hydrogen and nitrogen: $$ 3 \mathrm{H}_{2}(\mathrm{g})+\mathrm{N}_{2}(\mathrm{g}) \rightarrow 2 \mathrm{NH}_{3}(\mathrm{g}) $$ (a) If you want to produce 562 g of \(\mathrm{NH}_{3}\), what volume of \(\mathrm{H}_{2}\) gas, at \(56^{\circ} \mathrm{C}\) and \(745 \mathrm{mm}\) Hg, is required? (b) Nitrogen for this reaction will be obtained from air. What volume of air, measured at \(29^{\circ} \mathrm{C}\) and \(745 \mathrm{mm}\) Hg pressure, will be required to provide the nitrogen needed to produce \(562 \mathrm{g}\) of \(\mathrm{NH}_{3} ?\) Assume the sample of air contains 78.1 mole \(\%\) N \(_{2}\)

A xenon fluoride can be prepared by heating a mixture of \(\mathrm{Xe}\) and \(\mathrm{F}_{2}\) gases to a high temperature in a pressureproof container. Assume that xenon gas was added to a \(0.25-\mathrm{L}\) container until its pressure reached \(0.12 \mathrm{atm}\) at \(0.0^{\circ} \mathrm{C} .\) Fluorine gas was then added until the total pressure reached 0.72 atm at \(0.0^{\circ} \mathrm{C}\). After the reaction was complete, the xenon was consumed completely, and the pressure of the \(\mathrm{F}_{2}\) remaining in the container was 0.36 atm at \(0.0^{\circ} \mathrm{C} .\) What is the empirical formula of the xenon fluoride?

Analysis of a gaseous chlorofluorocarbon, \(\mathrm{CCl}_{x} \mathrm{F}_{y}\), shows that it contains \(11.79 \%\) C and \(69.57 \%\) Cl. In another experiment, you find that \(0.107 \mathrm{g}\) of the compound fills a 458 -mL. flask at \(25^{\circ} \mathrm{C}\) with a pressure of \(21.3 \mathrm{mm}\) Hg. What is the molecular formula of the compound?

The ideal gas law is least accurate under conditions of high pressure and low temperature. In those situations, using the van der Waals equation is advisable. (a) Calculate the pressure exerted by \(12.0 \mathrm{g}\) of \(\mathrm{CO}_{2}\) in a \(500-\mathrm{mL}\) vessel at \(298 \mathrm{K},\) using the ideal gas equation. Then, recalculate the pressure using the van der Waals equation. Assuming the pressure calculated from van der Waal's equation is correct, what is the percent error in the answer when using the ideal gas equation? (b) Next, cool this sample to \(-70^{\circ} \mathrm{C}\). Then perform the same calculation for the pressure exerted by \(\mathrm{CO}_{2}\) at this new temperature, using both the ideal gas law and the van der Waals equation. Again, what is the percent error when using the ideal gas equation?

A helium-filled balloon of the type used in longdistance flying contains \(420,000 \mathrm{ft}^{3}\left(1.2 \times 10^{7} \mathrm{L}\right)\) of helium. Suppose you fill the balloon with helium on the ground, where the pressure is 737 mm \(\mathrm{Hg}\) and the temperature is \(16.0^{\circ} \mathrm{C}\). When the balloon ascends to a height of 2 miles, where the pressure is only \(600 .\) mm Hg and the temperature is \(-33^{\circ} \mathrm{C},\) what volume is occupied by the helium gas? Assume the pressure inside the balloon matches the external pressure. Comment on the result.
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