Ncert Exercises & Solutions

We have 0.5 g of hydrogen gas in a cubic chamber of size 3 cm kept at NTP. The gas in the chamber is compressed keeping the temperature constant till a final pressure of 100 atm. Is one justified in assuming the ideal gas law, in the final state? (Hydrogen molecules can be considered as spheres of radius 1 $o A$ $\overset{o}{A}$ ).

Hint: Use the ideal gas equation.

Step 1: Find the volume of hydrogen molecules.

Assuming hydrogen molecules as spheres of radius 1

o A

$\overset{o}{A}$ .
So, the radius, r=1

o A

$\overset{o}{A}$
The volume of hydrogen molecules

= \frac{4}{3} {πr}^{3}

$= \frac{4}{3} {πr}^{3}$

= \frac{4}{3} (3.14) {(10^{- 10})}^{3}

$= \frac{4}{3} (3.14) {(10^{- 10})}^{3}$

= 4 \times 10^{- 30} m^{3}

$= 4 \times 10^{- 30} m^{3}$

Number of moles of

H_{2} = \frac{M a s s}{M o l e c u l a r m a s s}

$H_{2} = \frac{M a s s}{M o l e c u l a r m a s s}$

= \frac{0.5}{2} = 0.25

$= \frac{0.5}{2} = 0.25$

Molecules of

H_{2}

$H_{2}$ present = Number of moles of

H_{2}

$H_{2}$ present

\times 6.023 \times 10^{23}

$\times 6.023 \times 10^{23}$

= 0.25 \times 6.023 \times 10^{23}

$= 0.25 \times 6.023 \times 10^{23}$

$∴$ The volume of molecules present = number of molecules x volume of each molecule

$= 0.25 \times 6.023 \times 10^{23} \times 4 \times 10^{- 30}$

$= 6.023 \times 10^{23} \times 10^{- 30}$

$= 6 \times 10^{- 7} m^{3}$ ...(i)

Step 2: Find the final volume of the gas.

Now, if the ideal gas law is considered to be followed,

$p_{i} V_{i} = p_{f} V_{f}$

$V_{f} = (\frac{p_{i}}{p_{f}}) V_{i} = (\frac{1}{100}) {(3 \times 10^{- 2})}^{3}$

$= \frac{27 \times 10^{- 6}}{10^{2}}$

$= 2.7 \times 10^{- 7} m^{3}$ ...(ii)

Hence, on compression, the volume of the gas is of the order of the molecular volume [form Eq. (i) and Eq. (ii)]. The intermolecular forces will play a role and the gas will deviate from ideal gas behaviour.

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