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Question: Modern vacuum pumps can evacuate a vessel down to a pressure of \(4.0 \times {10^{ - 15}}\,atm\) at ...

Modern vacuum pumps can evacuate a vessel down to a pressure of 4.0×1015atm4.0 \times {10^{ - 15}}\,atm at room temperature (300K)\left( {300\,K} \right). Taking R=8.3KJmole1R = 8.3\,KJmol{e^{ - 1}} , 1atm=105Pa1\,atm = {10^5}\,Pa and Navogadro=6×1023mole1{N_{avogadro}} = 6 \times {10^{23}}\,mol{e^{ - 1}} , the mean distance between the molecules of gas in an evacuated vessel will be of the order of
A. 0.2μm0.2\,\mu m
B. 0.2mm0.2\,mm
C. 0.2cm0.2\,cm
D. 0.2nm0.2\,nm

Explanation

Solution

Here we have to use the concept of ideal gas laws. The ideal gas law is the equation of the state of a hypothetical ideal gas, sometimes called the general gas equation. In many conditions, it is a reasonable approximation of the behaviour of several gases, but it has many drawbacks.

Complete step by step answer:
The Ideal Gas Equation in the form PV=nRTPV = nRT is an excellent method for understanding the relationship in a given environment between the pressure, volume, quantity and temperature of an ideal gas that can be regulated for constant volume.
The ideal gas law can be derived from universal principles, but was originally derived from Charles's experimental measurements (that the volume filled by the gas is proportional to the temperature at a fixed pressure) and from Boyle's law (that the PVPV product is a constant at a fixed temperature). The volume filled by its atoms and molecules is a negligible fraction of VV in the ideal gas model. Under most conditions, the ideal gas law defines the behaviour of real gases.

Given,
Pressure, P=4.0×1015atmP = 4.0 \times {10^{ - 15}}\,atm, R=8.3KJmole1R = 8.3\,KJmol{e^{ - 1}}, 1atm=105Pa1\,atm = {10^5}\,Pa
Navogadro=6×1023mole1{N_{avogadro}} = 6 \times {10^{23}}\,mol{e^{ - 1}}. Using the relation PV=NkTPV = NkT, we get:
N=PkT N=4×1015×1.01×1058.3146.023×1023×300 N=1×105cm3  N = \dfrac{P}{{kT}} \\\ \Rightarrow N = \dfrac{{4 \times {{10}^{ - 15}} \times 1.01 \times {{10}^5}}}{{\dfrac{{8.314}}{{6.023 \times {{10}^{23}}}} \times 300}} \\\ \Rightarrow N = 1 \times {10^5}\,c{m^{ - 3}} \\\
The number of moles in 1cm=1051\,cm = {10^5}. So, one mole contains a volume of 105cm3{10^{ - 5}}\,c{m^3}.
For cube with side ll the volume will be
v13=1053 v13=0.2mm{v^{\dfrac{1}{3}}} = {10^{\dfrac{{ - 5}}{3}}} \\\ \therefore{v^{\dfrac{1}{3}}}= 0.2\,mm

Hence option B is the correct answer.

Note: Here we have to mainly notice what the units are. If the units are wrongly put in confusion, then the answers with powers would be wrong.The ideal gas law is closely related to energy: joules are the units on both sides. In PV=NkTPV = NkT, the right-hand side of the ideal gas rule is NkTNkT . This concept is approximately the sum of translational kinetic energy at the absolute temperature TT of NN atoms or molecules. PVPV, which also has the Joules units, is the left-hand side of the ideal gas law.