91Ó°ÊÓ

Pressure is defined as the force applied perpendicular to an object's surface per unit area across which that force is diffused. Gauge pressure is the pressure measured in proportion to the ambient pressure. Pressure may be measured in a number of different ways.

We have

\({\rm{PV = nRT}}\)

where,

\({\rm{P = }}\)pressure of the gas.

\({\rm{V = }}\)volume of the gas.

\({\rm{n = }}\)number of moles of the gas.

\({\rm{T = }}\)temperature of the gas.

Ifthe amount of the gas\({\rm{(n)}}\)is in moles (mol),temperature\({\rm{(T)}}\)is in kelvin (K),and pressure\({\rm{(P)}}\)is in atmospheres (atm).

\(\begin{aligned}{}\frac{{\rm{P}}}{{{\rm{RT}}}}{\rm{ = }}\frac{{\rm{n}}}{{\rm{V}}}\\{\rm{P = }}\frac{{{\rm{nRT}}}}{{\rm{V}}}\end{aligned}\)

We are given that

\(\begin{aligned}{}{\rm{Volume (V) = 36L}}{\rm{.}}\\{\rm{Pressure (P) = ? atm}}{\rm{.}}\\{\rm{Temperature (T) = 2}}{{\rm{5}}^{\rm{^\circ }}}{\rm{C}}{\rm{.}}\end{aligned}\)

Converting the temperatures in degree Celsius to kelvin, we have

\({\rm{ We have, }}{{\rm{0}}^{\rm{^\circ }}}{\rm{C = (0) + 273}}{\rm{.15K}}\)

So,\({\rm{2}}{{\rm{5}}^{\rm{^\circ }}}{\rm{C = (25) + 273}}{\rm{.15K = 298}}{\rm{.15K}}\)

Calculating the number of moles of\({\rm{C}}{{\rm{O}}_{\rm{2}}}\):

Atomic weight of Carbon\({\rm{(C) = 12}}\)g

Atomic weight of\({\rm{Oxygen(O) = 16}}\)g.

So, molecular weight of\({\rm{C}}{{\rm{O}}_{\rm{2}}}{\rm{ = 12 + (2) \times (16)}}\)g

\(\begin{aligned}{l}{\rm{ = 12 + 32}}\\{\rm{ = 44 grams}}{\rm{.}}\end{aligned}\)

Molar mass of\({\rm{C}}{{\rm{O}}_{\rm{2}}}{\rm{ = 44}}\frac{{{\rm{ gram }}}}{{{\rm{ mol }}}}\).

i.e.,\({\rm{1}}\)mole of\({\rm{C}}{{\rm{O}}_{\rm{2}}}{\rm{ = 44}}\)grams.

\({{\rm{n}}_{\rm{1}}}\)moles of\({\rm{C}}{{\rm{O}}_{\rm{2}}}{\rm{ = 350}}\)grams.

\({{\rm{n}}_{\rm{1}}}{\rm{ = }}\frac{{{\rm{350}}}}{{{\rm{44}}}}{\rm{ = 7}}{\rm{.9 moles of C}}{{\rm{O}}_{\rm{2}}}\)

Calculating the number of moles of\({{\rm{O}}_{\rm{2}}}\):

Atomic weight of Oxygen\({\rm{(O) = 16}}\)gram.

So, molecular weight of\({{\rm{O}}_{\rm{2}}}{\rm{ = (2) \times (16)}}\)grams.\({\rm{ = 32}}\)grams.

Molar mass of\({{\rm{O}}_{\rm{2}}}{\rm{ = 32}}\frac{{{\rm{ gram }}}}{{{\rm{mol}}}}\).

i.e.\({\rm{1}}\)mole of\({{\rm{O}}_{\rm{2}}}{\rm{ = 32}}\)grams.

\({{\rm{n}}_{\rm{2}}}\)moles of\({{\rm{O}}_{\rm{2}}}{\rm{ = 805}}\)grams.

\({{\rm{n}}_{\rm{2}}}{\rm{ = }}\frac{{{\rm{805}}}}{{{\rm{32}}}}\)

\({\rm{ = 25}}{\rm{.1}}\)moles of\({{\rm{O}}_{\rm{2}}}\)

Calculating no. of moles of\({{\rm{N}}_{\rm{2}}}\)

Atomic weight of Nitrogen\((N) = 14\)gram.

So, molecular weight of\({{\rm{N}}_{\rm{2}}}{\rm{ = (2) \times (14)}}\)grams.\({\rm{ = 28}}\)grams.

Molar mass of\({{\rm{N}}_{\rm{2}}}{\rm{ = 28}}\frac{{{\rm{gram}}}}{{{\rm{mol}}}}{\rm{.}}\)

i.e.\({\rm{1}}\)mole of\({{\rm{N}}_{\rm{2}}}{\rm{ = 32}}\)grams.

\({{\rm{n}}_{\rm{3}}}\)moles of\({{\rm{N}}_{\rm{2}}}{\rm{ = 4880}}\)grams.

\({{\rm{n}}_{\rm{3}}}{\rm{ = }}\frac{{{\rm{4880}}}}{{{\rm{28}}}}{\rm{ = 174}}{\rm{.2 moles of }}{{\rm{N}}_{\rm{2}}}\)

So, we have the ratio as

\({\rm{7}}{\rm{.9C}}{{\rm{O}}_{\rm{2}}}{\rm{:25}}{\rm{.1}}{{\rm{O}}_{\rm{2}}}{\rm{:174}}{\rm{.2\;}}{{\rm{N}}_{\rm{2}}}\)

Total number of moles of gas inside the container\({\rm{(n) = }}{{\rm{n}}_{\rm{1}}}{\rm{ + }}{{\rm{n}}_{\rm{2}}}{\rm{ + }}{{\rm{n}}_{\rm{3}}}\)

\({\rm{ = 7}}{\rm{.9 + 25}}{\rm{.1 + 174}}{\rm{.2}}\)

\({\rm{ = 207}}{\rm{.2}}\)moles

\({\rm{PV = nRT}}\)

\(\begin{aligned}{}{\rm{Pressure (P) = }}\frac{{{\rm{nRT}}}}{{\rm{V}}}\\{\rm{ = }}\frac{{{\rm{(207}}{\rm{.2) \times (0}}{\rm{.08206) \times (298}}{\rm{.15)}}}}{{{\rm{36}}}}\\{\rm{ = }}\frac{{{\rm{5069}}{\rm{.4}}}}{{{\rm{36}}}}\\{\rm{ = 141\;atm}}{\rm{.}}\end{aligned}\)

Therefore, the pressure is \({\rm{141\;atm}}{\rm{.}}\)