States of Matter Solid, Liquid and Gas MCQ Questions & Answers in Physical Chemistry | Chemistry

\[\underset{\begin{smallmatrix} \\ \text{Initial mole} \\ \text{Final mole} \end{smallmatrix}}{\mathop{{}}}\,\underset{\begin{smallmatrix} 2a \\ \left( 2a-2x \right) \end{smallmatrix}}{\mathop{2{{H}_{2\left( g \right)}}}}\,+\underset{\begin{smallmatrix} a \\ \left( a-x \right) \end{smallmatrix}}{\mathop{{{O}_{2\left( g \right)}}}}\,\to \underset{\begin{smallmatrix} 0 \\ 2x \end{smallmatrix}}{\mathop{2{{H}_{2}}{{O}_{\left( g \right)}}}}\,\]
$$\eqalign{ & {\text{Given,}}\,\,2x = 2a \times \frac{{80}}{{100}} = 1.6a \cr & \therefore \,\,x = 0.8a \cr} $$
$${\text{Thus, after the reaction}}\,\,{H_2}\,\,{\text{left}}$$       $$ = 2a - 1.6a = 0.4a\,\,mole$$
$$\eqalign{ & {O_2}\,\,{\text{left}} = 0.2a\,\,mole \cr & {H_2}O\,\,{\text{formed}} = 1.6a\,\,mole \cr & \therefore \,\,{\text{Total mole at}}\,\,120{\,^ \circ }C\,\,{\text{in gaseous phase}} \cr & = 0.4a + 0.2a + 1.6a = 2.2a \cr} $$
$${\text{Now, at initial conditions}}\,\,$$     $$P = 0.8\,atm,T = 293\,K$$
$$\eqalign{ & P \times V = nRT \cr & 0.8 \times V = 3a \times R \times 293 \cr & \therefore \,\,V = \frac{{3a \times R \times 293}}{{0.8}} \cr} $$
$${\text{The volume of container}}$$     $${\text{remains constant}}{\text{.}}$$
$${\text{Using,}}\,\,P \times V = nRT$$
$$\therefore \,\,P \times \frac{{3a \times R \times 293}}{{0.8}} = 2.2a \times R \times 393$$           $$\left[ {T = 393\,K} \right]$$
$$\eqalign{ & P = \frac{{393 \times 0.8 \times 2.2}}{{3 \times 293}} \cr & \,\,\,\,\,\, = 0.787\,atm \cr} $$

At high altitudes, atmospheric pressure is low, hence water boils at a lower temperature.

For an ideal-gas behaviour, the molecules of a gas should be far apart. The factors favouring this condition are high temperature and low pressure.

$$\eqalign{ & {\text{Molar mass}}\, \uparrow ,\,\,'a'\,\,{\text{increases}} \cr & {\text{size of molecule}}\, \uparrow ,\,'b'\,{\text{increase}} \cr & b\left( {L/mol} \right)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,a\left( {bar.\,{L^2}/mo{l^2}} \right) \cr & {H_2} \to 0.02661\,\,\,\,\,\,\,\,\,\,\,\,\,C{H_4} \to 2.25 \cr & He \to 0.0237\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{O_2} \to 1.36 \cr & {O_2} \to 0.03183\,\,\,\,\,\,\,\,\,\,\,\,\,\,{H_2} \to 0.244 \cr & C{O_2} \to 0.04267 \cr} $$

In case of face centred cubic $$(fcc)$$  lattice,
radius $$ = \frac{{\sqrt 2 a}}{4}$$
∴ Radius of copper atom ( $$fcc$$  lattice )
$$\eqalign{ & = \frac{{\sqrt 2 \times 361}}{4} \cr & = 128\,pm \cr} $$

$$\eqalign{ & {\text{The mean free path,}} \cr & \lambda = \frac{1}{{\sqrt 2 \pi {a^2}N}} \cr & {\text{or}}\,\lambda \alpha \frac{1}{{{a^2}}},\,{\text{where}}\,a = {\text{molecular}}\,{\text{diameter}} \cr & \therefore \,\,\,{\text{Smaller the molecular diameter, longer the mean free}} \cr & {\text{path}}{\text{. Hence}}\,{H_2}\,{\text{is}}\,{\text{the}}\,{\text{answer}}{\text{.}} \cr} $$

$$KE = \frac{3}{2}RT$$   ( for one mole of a gas )
As, the kinetic energy of a gaseous molecule depends only on temperature, thus at constant temperature, the kinetic energy of the molecules remains the same.

$$\eqalign{ & {\text{Compressibility factor}}\,\left( Z \right) = \frac{{PV}}{{RT}} \cr & {\text{(For one mole of real gas)}} \cr & {\text{van der Waals equation}} \cr & \left( {P + \frac{a}{{{V^2}}}} \right)\left( {V - b} \right) = RT \cr} $$
At low pressure, volume is very large and hence correction term $$b$$ can be neglected in comparison to very large volume of $$V.$$
$$\eqalign{ & {\text{i}}{\text{.e}}{\text{.}}\,\,V - b \approx V \cr & \left( {P + \frac{a}{{{V^2}}}} \right)V = RT\,;\,PV + \frac{a}{V} = RT \cr & PV = RT - \frac{a}{V};\,\frac{{PV}}{{RT}} = 1 - \frac{a}{{VRT}} \cr} $$
$${\text{Hence}},$$ 

Kinetic theory of gases proves all the given gas laws.

Let the total mass of the mixture be $$100\,g$$
Mass of $${H_2} = \frac{{20}}{{100}} \times 100 = 20\,g$$
No. of moles of $${H_2} = \frac{{20}}{2} = 10\,moles$$
Mass of $${O_2} = 100 - 20 = 80\,g$$
No. of moles of $${O_2} = \frac{{80}}{{32}} = 2.5\,moles$$
Total number of moles of $${H_2}$$  and $${O_2} = 10 + 2.5 = 12.5$$
Mole fraction of $${H_2} = \frac{{10}}{{12.5}} = 0.8$$
Total pressure of mixture $$ = 1\,bar$$
∴ Partial pressure of $${H_2}$$  in the mixture, $${p_{{H_2}}}$$
$$\eqalign{ & = {\text{Mole fraction }} \times {\text{ Total pressure}} \cr & = 0.8 \times 1 \cr & = 0.8\,bar \cr} $$