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

$$\eqalign{ & PV = {\text{constant at given temperature}} \cr & \therefore \,\,{P_1}{V_1} = {P_2}{V_2} = {P_3}{V_3} = {P_4}{V_4} \cr} $$
$${\text{Now,}}\,\,{V_1} > {V_2} > {V_3} > {V_4}$$         $$\left( {{\text{From the figure}}} \right)$$
$${\text{Hence,}}\,\,{P_1} < {P_2} < {P_3} < {P_4}$$         $$\left( {\because \,\,P \propto \frac{1}{V}} \right)$$

Let the mass of $${H_2}$$ gas be $$xg$$  and mass $${O_2}$$ gas $$4xg$$
$$\eqalign{ & Molar\,\,\,\,{H_2}\,\,:\,\,\,{O_2}\,\, \cr & mass\,\,\,\,\,\,\,\,2\,\,\,\,:\,\,\,32 \cr & {\text{i}}{\text{.e}}{\text{.}}\,\,\,\,\,\,\,\,\,\,\,\,1\,\,\,\,\,\,:\,\,\,16 \cr & \therefore \,\,{\text{Molar ratio}} = \frac{{{n_{{H_2}}}}}{{{n_{{O_2}}}}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{\frac{x}{2}}}{{\frac{{4x}}{{32}}}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{x \times 32}}{{2 \times 4x}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{4}{1} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 4:1 \cr} $$

In a face centred cubic $$(fcc)$$  lattice, a unit cell is shared equally by six unit cells.

$$\eqalign{ & PV = nRT \cr & \Rightarrow P = \frac{{4 \times 0.0821 \times 300}}{{89.6}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\, = 1.099\,atm \cr} $$

Partial charges are always less than the unit electronic charge $$\left( {1.6 \times {{10}^{ - 19}}C} \right).$$

$$\eqalign{ & PV = nRT\,\,{\text{or}}\,\,T = \frac{{PV}}{{nR}}\,\,\,\,\left( {\because \,\,n = \frac{m}{M}} \right) \cr & T = \frac{{2 \times 20 \times 28}}{{28 \times 0.0821}} = 487.2\,K \cr} $$

$$\eqalign{ & T{V^{\gamma - 1}} = {\text{Constant}}\,\,\,\,\,\,\,\left( {\because \,\,{\text{Change}}\,{\text{is}}\,{\text{adiabatic}}} \right) \cr & {T_1}{V_1}^{^{\gamma - 1}} = {T_2}V_2^{^{\gamma - 1}} \cr & {\text{For monoatomic gas}}\,\,\gamma = \frac{5}{3} \cr & \therefore \,\,{T_1}V_1^{\frac{2}{3}} = {T_2}V_2^{\frac{2}{3}} \Rightarrow T{\left( 1 \right)^{\frac{2}{3}}} = {T_2}{\left( 2 \right)^{\frac{2}{3}}} \cr & {T_2} = \frac{T}{{{2^{\left( {\frac{2}{3}} \right)}}}} \cr} $$

$$\frac{{{r_1}}}{{{r_2}}} = \sqrt {\frac{{{d_2}}}{{{d_1}}}} = \sqrt {\frac{{{M_2}}}{{{M_1}}}} $$

According to Avogadro's hypothesis,
Volume of a gas $$(V) \propto $$   number of moles $$(n)$$
Therefore, the ratio of the volumes of gases can be determined in terms of their moles.
∴ The ratio of volumes of $${H_2}:{O_2}:{\text{methane}}\left( {C{H_4}} \right)$$     is given by
$$\eqalign{ & {V_{{H_2}}}:{V_{{O_2}}}:{V_{C{H_4}}} = {n_{{H_2}}}:{n_{{O_2}}}:{n_{C{H_4}}} \cr & \Rightarrow {V_{{H_2}}}:{V_{{O_2}}}:{V_{C{H_4}}}:\, = \frac{{{m_{{H_2}}}}}{{{M_{{H_2}}}}}:\frac{{{m_{{O_2}}}}}{{{M_{{O_2}}}}}:\frac{{{m_{C{H_4}}}}}{{{M_{C{H_4}}}}} \cr} $$
$${\text{Given,}}$$   $${m_{{H_2}}} = {m_{{O_2}}} = {m_{C{H_4}}} = m$$     $$\left[ {\because \,n = \frac{{{\text{mass}}}}{{{\text{molar}}\,{\text{mass}}}}} \right]$$
$${\text{Thus}},$$   $${V_{{H_2}}}:{V_{{O_2}}}:{V_{C{H_4}}} = \frac{m}{2}:\frac{m}{{32}}:\frac{m}{{16}}$$       $$ = 16:1:2$$

$$\eqalign{ & {\text{We know that ideal gas equation is}} \cr & pV = nRT \cr & pV = \frac{w}{M}RT \cr & pM = \frac{w}{V}RT\,\,\,\,\left[ {\because \,\,\frac{w}{V} = d} \right] \cr & pM = dRT \cr & \Rightarrow d = \frac{{pM}}{{RT}} \cr} $$