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 $\overset{o}{A}$ ).

Hint: Use the ideal gas equation.

Step 1: Find the volume of hydrogen molecules.

Assuming hydrogen molecules as spheres of radius 1

\overset{o}{A}

.
So, the radius, r=1

\overset{o}{A}

The volume of hydrogen molecules

= \frac{4}{3} {πr}^{3}

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

= \frac{0.5}{2} = 0.25

Molecules of

H_{2}

present = Number of moles of

H_{2}

present

\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.

NEET Information

courses

Company

Botany Questions

Chemistry Questions

Physics Questions

Zoology Questions