The best vacuum that can be attained in the laboratory corresponds to a pressure of about \(10^{-18} \mathrm{~atm}\), or \(1.01 \times 10^{-13} \mathrm{~Pa}\). How many molecules are there per cubic centimeter in such a vacuum at \(22^{\circ} \mathrm{C} ?\)

Short Answer

Expert verified
The number of molecules per cubic centimeter in such a vacuum at \(22^{\circ} \mathrm{C}\) is given by the formula \(n \times N_A\). To find the exact numerical value, one would have to compute the calculations in Steps 2 and 3.

Step by step solution

01

Convert temperature to Kelvin

To get the temperature in Kelvin, add 273 to the temperature in degrees Celsius. \(T = 22^{\circ} \mathrm{C} + 273 = 295 \, K\)
02

Apply the Ideal Gas Law

Rearrange the Ideal Gas Law to solve for n (number of moles), so \(n = PV/RT\). Substitute the given values \(P = 1.01 \times 10^{-13} \mathrm{Pa}\), \(V = 1 \mathrm{cm^{3}} = 1 \times 10^{-6} \mathrm{m^3}\), \(R = 8.31 \mathrm{J/mol \cdot K}\) and \(T = 295 \mathrm{K}\) to get \(n = PV/RT = 1.01 \times 10^{-13} \mathrm{Pa} \cdot 1 \times 10^{-6} \mathrm{m^{3}} / (8.31 \mathrm{J/mol \cdot K} \cdot 295 \mathrm{K})\)
03

Convert moles to molecules

Multiply the volume in moles by Avogadro's number (\(6.02 \times 10^{23} \) molecules/mole) to obtain the number of molecules: \(n \times N_{A}\), where \(N_A\) is Avogadro's number.

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Key Concepts

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

Molecules per Cubic Centimeter
Understanding the concept of molecules per cubic centimeter involves diving into the microscopic world of gases. A cubic centimeter is a tiny cube that is 1 cm long on each side. The number of molecules in this tiny volume can vary vastly depending on the gas and its conditions. To visualize this concept, imagine a tiny box filled with an incredible number of tiny particles, all moving around and colliding with each other.

To find the number of molecules in such a space, scientists use measurements like pressure and temperature, combined with constants and equations from gas laws. In our exercise, we're given an exceedingly low pressure, typical of an ultra-high vacuum, where particles are extremely sparse, reflecting the sheer vastness of empty space even in a small volume.
Gas Pressure Calculation
Gas pressure is an expression of the force exerted by gas molecules as they collide with the walls of their container. Calculating the pressure of a gas requires knowledge of several variables such as volume, temperature, and the amount of gas present, reflected in the ideal gas law, expressed as PV=nRT. This equation is powerful because it allows us to understand the behaviour of gases under different conditions.

In the laboratory vacuum example, the pressure is extremely low, reflecting the scarce number of molecular collisions within the cubic centimeter volume. By rearranging the ideal gas law, we're able to isolate and solve for the number of moles, and hence determine the gas pressure for given temperature and volume conditions.
Avogadro's Number
Avogadro's number is a fundamental constant in chemistry, pegged at approximately 6.02 x 10^23. This large number signifies how many particles—which could be atoms or molecules—exist in one mole of a substance. The mole is a standard unit in chemistry that allows scientists to count particles by weighing them, connecting the microscopic scale to the lab-scale.

When we talk about molecules per cubic centimeter, we tie this vast number into our calculations by converting moles—determined from the ideal gas law—to a count of molecules using Avogadro's number. This step provides a staggering perspective on the number of particles playing their part in what seems like insignificant amounts of matter.
Converting Celsius to Kelvin
Temperature scales can be confusing, as different scales are used for different applications. Celsius is commonly used in daily life and some scientific work, but Kelvin is the standard unit of temperature in the physical sciences. To convert Celsius to Kelvin, we add 273.15 to the Celsius temperature, because 0 degrees Celsius corresponds to 273.15 Kelvin, the temperature at which water freezes under standard atmospheric conditions.

In our example, we convert 22 degrees Celsius to Kelvin by adding 273. This step is necessary for gas calculations because the ideal gas law requires temperature to be in Kelvin. This conversion ensures that our temperature scale has no negative values, aligning with the absolute thermodynamic temperature scale where zero Kelvin is absolute zero, the coldest possible temperature where molecules theoretically stop moving.

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Most popular questions from this chapter

A rod is measured to be \(20.05 \mathrm{~cm}\) long using a steel ruler at a room temperature of \(20^{\circ} \mathrm{C}\). Both the rod and the ruler are placed in an oven at \(270^{\circ} \mathrm{C}\), where the rod now measures \(20.11 \mathrm{~cm}\) using the same rule. Calculate the coefficient of thermal expansion for the material of which the rod is made.

At what temperature is the Fahrenheit scale reading equal to (a) twice that of the Celsius and \((b)\) half that of the Celsius?

An aluminum cup of \(110 \mathrm{~cm}^{3}\) capacity is filled with glycerin at \(22^{\circ} \mathrm{C}\). How much glycerin, if any, will spill out of the cup if the temperature of the cup and glycerin is raised to \(28^{\circ} \mathrm{C}\) ? (The coefficient of volume expansion of glycerin is \(5.1 \times\) \(\left.10^{-4} / \mathrm{C}^{\circ} .\right)\)

Oxygen gas having a volume of \(1130 \mathrm{~cm}^{3}\) at \(42.0^{\circ} \mathrm{C}\) and a pressure of \(101 \mathrm{kPa}\) expands until its volume is \(1530 \mathrm{~cm}^{3}\) and its pressure is \(106 \mathrm{kPa}\). Find \((a)\) the number of moles of oxygen in the system and ( \(b\) ) its final temperature.

At \(100^{\circ} \mathrm{C}\) a glass flask is completely filled by \(891 \mathrm{~g}\) of mercury. What mass of mercury is needed to fill the flask at \(-35^{\circ} \mathrm{C} ?\) (The coefficient of linear expansion of glass is \(9.0 \times 10^{-6} / \mathrm{C}^{\circ} ;\) the coefficient of volume expansion of mercury is \(1.8 \times 10^{-4} / \mathrm{C}^{\circ}\).)

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