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

$$N{H_3}$$  has higher value of $$a,$$  it means intermolecular forces are stronger and higher critical temperature than oxygen, hence easy to liquefy.

At any given pressure, graph of volume vs temperature ( in $$^ \circ C$$ ) is a straight line and on extending to zero volume each line intercepts the temperature axis at $$ - {273.15^ \circ }C.$$

Let the mass of methane and oxygen $$ = \,m\,gm.$$
Mole fraction of $${O_2}$$
$$\eqalign{ & = \frac{{{\text{Moles of}}\,{O_2}}}{{{\text{Moles of}}\,{O_2} + {\text{Moles of}}\,C{H_4}}} \cr & = \frac{{\frac{m}{{32}}}}{{\frac{m}{{32}} + \frac{m}{{16}}}} \cr & = \frac{{\frac{m}{{32}}}}{{\frac{{3m}}{{32}}}} \cr & = \frac{1}{3} \cr} $$
Partial pressure of $${O_2} = $$  Total pressure × mole fraction of $${O_2},$$
$$\eqalign{ & {P_{{O_2}}} = P \times \frac{1}{3} \cr & = \frac{1}{3}P \cr} $$

Liquids and gases, both can diffuse and take the shape of the containers.

Kinetic energy of a gas depends upon the temperature. At constant temperature, kinetic energy of all gas molecules is same.

A solution of $$C{H_3}OH$$   and water shows positive deviation from Raoult's law, it means by adding $$C{H_3}OH$$  intermolecular force of attraction decreases and hence surface tension decreases.
$$ \bullet $$ By adding $$KCl$$  in water, intermolecular force of attraction bit increases, so surface tension increases by small value.
$$ \bullet $$ By adding surfactant like $$C{H_3}{\left( {C{H_2}} \right)_{11}}OSO_3^ - N{a^ + },$$     surface tension decreases rapidly and after forming micelle it slightly increases.

The van der Waals’ equation is
$$\left( {p + \frac{a}{{{V^2}}}} \right)\left( {V - b} \right) = RT$$
The term $$\left( {p + \frac{a}{{{V^2}}}} \right)$$   used for pressure correction, it measures the intermolecular forces between the molecules of gas.

Boiling point of a Iiquid is the temperature at which vapour pressure becomes equal to the atmospheric pressure. Lower the atmospheric pressure, lower is the boiling point.

$$\eqalign{ & {\text{For the reaction,}} \cr & SrC{O_3}\left( s \right) \rightleftharpoons SrO\left( s \right) + C{O_2}\left( q \right), \cr} $$
$${K_p} = 1.6\,atm = {p_{C{O_2}}}$$     $$ = {\text{maximum pressure of}}\,\,C{O_2}$$
$${\text{Given,}}\,{p_1} = 0.4\,atm,\,{V_1} = 20\,L,$$       $${T_1} = 400\,K$$
$${p_2} = 1.6\,atm,\,{V_2} = ?,\,{T_2} = 400\,K$$
$$\eqalign{ & {\text{At constant temperature,}} \cr & \,\,\,\,\,\,\,\,\,{p_1}{V_1} = {p_2}{V_2} \cr & 0.4 \times 20 = 1.6 \times {V_2} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,{V_2} = \frac{{0.4 \times 20}}{{1.6}} = 5\,L \cr} $$

$$PV = nRT = \frac{m}{M}RT$$
$$P = 5\,atm,V = 10\,L,T = 300\,K,$$       $$R = 0.0821\,\,L\,\,atm\,\,{K^{ - 1}}\,\,mo{l^{ - 1}}$$
$$\eqalign{ & m = \frac{{PVM}}{{RT}} \cr & \,\,\,\,\,\, = \frac{{5 \times 10 \times 44}}{{0.0821 \times 300}} \cr & \,\,\,\,\,\, = 89.3\,g \cr} $$