Question

$\red{ \bold{ \underbrace{ \overbrace{ \underline{ \orange{Urgent \: Help \: Request!!!}}}}}}$
[Chemistry]

Calculate the volume of dry air at S.T.P. that occupies 28 cm3 at 14°C and 750 mm Hg pressure when saturated with water vapour. The vapour pressure of water at 14°C is 12 mm Hg.
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Answer 1

$\\ \large \underline \bold{ \: Question :-}$

Calculate the volume of dry air at S.T.P. that occupies 28 cm3 at 14°C and 750 mm Hg pressure when saturated with water vapour. The vapour pressure of water at 14°C is 12 mm Hg.

$\\ \large \underbrace{ \underline\bold{ \: Given :-}}$

$\\ \sf \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \bigstar \: \: P_2 = 760 \: mmHg$

$\\ \sf \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \bigstar \: \: P_1 = 750 - 12 = 738 \: mmHg$

$\\ \sf \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \bigstar \: \: T_1 = 14 \degree \:C = 14 + 273\: = \: 287 \: K$

$\\ \sf \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \bigstar \: \: T_2 = 0\degree \:C = 0+ 273\: = \: 273\: K$

$\\ \sf \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \bigstar \: \:V_1 = 28 \: \: {cm}^{3}$

$\\ \sf \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \bigstar \: \:V_2 = ?$

$\\\large\underbrace{ \underline\bold{ \: Using \: formula \ :-}}$

$\\ \sf \: \: \: \: \: \: \: \: \: \: \: \: \: \frac{P_1V_1}{T_1 }= \frac{P_2V_2}{T_2}$

$\\ \large \underbrace{ \underline\bold{ \: Solution :-}}$

Let put the values in above formula, We get

$\\ \sf \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \implies \: \frac{728\times38}{287}= \frac{760 × V_2}{273}$

$\\ \sf \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \implies \: V_2 = \frac{728\times38\times273}{287\times260}$

$\\ \sf \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \implies \: V_2 = \frac{5641272}{218120}$

$\\ \sf \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \implies \underline{ \boxed{\orange{ \frak{ V_2 = 25.86 ml}} }}$

Answer 2

Pressure due to dry air,

$\rm{P_{1} = 750 - 12 = 738 mm}$

$\rm{V_{1}  = {28 cm}^{2}}$

$\rm{T_{1} = {14}^{\circ}C = 14 + 273 = 287 K}$

‎

$\rm{P_{2} = 760 mmHg}$

$\rm{V_{2}  = ?}$

$\rm{T_{2} = {0}^{\circ}C = 273 K}$

‎

$\sf{\underline{Using\:gas\:law}}$

${\boxed{\dfrac{{P}_{1}{V}_{1}}{{T}_{1}} = {\dfrac{{P}_{2}{V}_{2}}{{T}_{2}}}}}$

‎

${\dfrac{738\:{\times}\:28}{287}} = \dfrac{760\:{\times}\:{V}_{2}}{273}$

${\cancel{\dfrac{20664}{287}}} = \dfrac{760\:{\times}\:{V}_{2}}{273}$

$72 = \dfrac{760\:{\times}\:{V}_{2}}{273}$

$72\:{\times}\:273 = 760\:{\times}\:{V}_{2}$

$19656 = 760\:{\times}\:{V}_{2}$

${V}_{2} = {\dfrac{19656}{760}}$

${V}_{2} = {\dfrac{{\cancel{2}^{3}}\:{\times}\:{3}^{3}\:{\times}\:7\:{\times}\:13}{{\cancel{2}^{3}}\:{\times}\:5\:{\times}\:19}}$

${V}_{2} = {\dfrac{{3}^{3}\:{\times}\:7\:{\times}\:13}{5\:{\times}\:19}}$

${V}_{2} = {\dfrac{27\:{\times}\:7\:{\times}\:13}{5\:{\times}\:19}}$

${\sf{\boxed{{V}_{2} = {\dfrac{2457}{95}}}\:{\approx}\:{\large{\boxed{\underline{\red{25.86\:ml}}}}}}}$