
In a very good vacuum system in the laboratory. The vacuum stained was ${10^{ - 13}}$ atm. If the temperature the system was \[300K\], the numbers of molecules present in a volume of $1c{m^3}$ is
A) ${\text{A}}{\text{. }}2.4 \times {10^6}$
B) ${\text{B}}{\text{. }}24$
C) ${\text{C}}{\text{. }}24 \times {10^9}$
D) ${\text{D}}{\text{.}}$ Zero
Answer
518.7k+ views
Hint:The empirical relationships among the volume, the temperature, the pressure, and the amount of a gas can be combined to the perfect gas equation.
Complete step by step:
For the general gas law we obtain,
$PV = nRT$
Where,
$p$ Is the pressure
$v$ Is the volume
$n$ Amount of Substance
$R$ Is the ideal gas constant
$T$ Is the absolute temperature.
$ \Rightarrow n = $ Number of moles present in the volume
$ \Rightarrow \dfrac{{PV}}{{RT}}$
The given values are applied by,
$ \Rightarrow \dfrac{{({{10}^{ - 13}} \times 1.01 \times {{10}^5}) \times {{({{10}^{ - 2}})}^2}}}{{8.31 \times 300}}$
$ \approx 4.05 \times {10^{ - 18}}$
Thus number of molecules in the given volume $ = n{N_A}$
${N_A} = 6.023 \times {10^{23}}$
$ \Rightarrow 4.05 \times {10^{ - 18}} \times 6.023 \times {10^{23}}$
$ \Rightarrow 2.4 \times {10^6}$
So, the correct answer is option \[\left( A \right)\]
Additional information:
A hypothetical gas composed of molecules which follow a few rules in the term of ideal gas refers. The molecules do not attract or repel in each other. The molecules would be elastic collisions upon impact with each other or an elastic collision with the walls of the container.
The ideal gas is three basic classes
The classical or Maxwell–Boltzmann ideal gas,
The ideal quantum are classified into Bose gas, composed of bosons, and the ideal quantum Fermi gas is composed by the Fermi distribution law.
Note:The following assumptions are depends on the ideal gas
Elastic to all collisions and is frictionless to all motion (no energy loss in motion or collision).
Apply by Newton’s laws.
The size of the molecules, the average distance between them.
They are constantly moving in random directions with a distribution of speeds of the molecules.
Complete step by step:
For the general gas law we obtain,
$PV = nRT$
Where,
$p$ Is the pressure
$v$ Is the volume
$n$ Amount of Substance
$R$ Is the ideal gas constant
$T$ Is the absolute temperature.
$ \Rightarrow n = $ Number of moles present in the volume
$ \Rightarrow \dfrac{{PV}}{{RT}}$
The given values are applied by,
$ \Rightarrow \dfrac{{({{10}^{ - 13}} \times 1.01 \times {{10}^5}) \times {{({{10}^{ - 2}})}^2}}}{{8.31 \times 300}}$
$ \approx 4.05 \times {10^{ - 18}}$
Thus number of molecules in the given volume $ = n{N_A}$
${N_A} = 6.023 \times {10^{23}}$
$ \Rightarrow 4.05 \times {10^{ - 18}} \times 6.023 \times {10^{23}}$
$ \Rightarrow 2.4 \times {10^6}$
So, the correct answer is option \[\left( A \right)\]
Additional information:
A hypothetical gas composed of molecules which follow a few rules in the term of ideal gas refers. The molecules do not attract or repel in each other. The molecules would be elastic collisions upon impact with each other or an elastic collision with the walls of the container.
The ideal gas is three basic classes
The classical or Maxwell–Boltzmann ideal gas,
The ideal quantum are classified into Bose gas, composed of bosons, and the ideal quantum Fermi gas is composed by the Fermi distribution law.
Note:The following assumptions are depends on the ideal gas
Elastic to all collisions and is frictionless to all motion (no energy loss in motion or collision).
Apply by Newton’s laws.
The size of the molecules, the average distance between them.
They are constantly moving in random directions with a distribution of speeds of the molecules.
Recently Updated Pages
Master Class 11 Economics: Engaging Questions & Answers for Success

Master Class 11 Accountancy: Engaging Questions & Answers for Success

Master Class 11 English: Engaging Questions & Answers for Success

Master Class 11 Social Science: Engaging Questions & Answers for Success

Master Class 11 Biology: Engaging Questions & Answers for Success

Master Class 11 Physics: Engaging Questions & Answers for Success

Trending doubts
1 ton equals to A 100 kg B 1000 kg C 10 kg D 10000 class 11 physics CBSE

Difference Between Prokaryotic Cells and Eukaryotic Cells

One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

1 Quintal is equal to a 110 kg b 10 kg c 100kg d 1000 class 11 physics CBSE

Proton was discovered by A Thomson B Rutherford C Chadwick class 11 chemistry CBSE

Draw a diagram of nephron and explain its structur class 11 biology CBSE
