Question

the lowest pressure that can be created in laboratory at 27 degree Celsius is
${10}^{ - 11}$
mmy of Hg. At this pressure, what is the number of ideal gas molecules per cm^3 ?

Answer 1

Answer: The number of ideal gas molecule per $cm^{3}$ is $\dfrac{n}{V} = 5.343\times10^{-19}mol/cm^{3}$.

Explanation:

Given that,

Temperature T = 27+273 = 300 K

Pressure $P = \dfrac{10^{-11}}{760}atm$

Gas constant R = 0.0821

Using ideal gas equation

$PV = nRT$

$\dfrac{n}{V}= \dfrac{P}{RT}$

$\dfrac{n}{V} = \dfrac{\dfrac{10^{-11}}{760}}{0.0821\times 300}$

$\dfrac{n}{V} = 5.343\times10^{-19}mol/cm^{3}$

Hence, the number of ideal gas molecule per $cm^{3}$ is $\dfrac{n}{V} = 5.343\times10^{-19}mol/cm^{3}$.

Answer 2

Given data:

Pressure is $P = 10^{11} \quad mm \quad of \quad Hg$ in which $Pascal, p\quad =\quad 1.33\times { 10 }^{ -9 }pa$

The temperature is stated as  which is Kelvin be T = 300 K.

Solution:

Therefore the ideal gas molecule can be given by – P V = n R T which can be in other terms will be P V = N K B T;  where $KB\quad =\quad 1.38\times { 10 }^{ -23 }$

$\Rightarrow \frac { N }{ V } \quad =\quad \frac { P }{ KBT }$

$\Rightarrow \frac { N }{ V } \quad =\quad \frac { 1.33\times { 10 }^{ -9 } }{ 1.38\times { 10 }^{ -3 } } \times 300$

$\Rightarrow \frac { N }{ V } \quad =\quad 3.22\times { 10 }^{ 11 }$

$T\quad=\quad 27 \quad degree \quad Celsius$