Question

a small bubble rise from a bottom of the sea where the temperature is 15°C and pressure of 8.2 atm reaches the surface of sea where the temperature is 30°C and pressure of 2 atm calculate the final volunce if the initial volunce is 2.4 cm^3​

Answer 1

Consider the Ideal Gas Equation,

$\tt PV = nRT \\ \\ \dashrightarrow \tt \dfrac{PV}{ T} = nR \\ \\ \dashrightarrow \tt \dfrac{PV}{ T} =constant$

Therefore,

$\dashrightarrow \boxed{ \boxed{ \tt \bigg( \dfrac{PV}{ T} \bigg)_1 = \tt \bigg(\dfrac{PV}{ T} \bigg)_2 }}$

Given Information,

• Initial Temperature = 15°C = 288 K
• Final Temprature = 30°C = 303 K
• Initial Volume = 2.4 cm³
• Final Volume (say V) = V cm³
• Initial Pressure =8.2 atm
• Final Pressure = 2 atm

Substituting the values,

$\tt \: \dfrac{8.2 \times 2.4}{288} = \dfrac{2 \times V}{303} \\ \\ \dashrightarrow \tt \: V = \dfrac{8.2 \times 2.4 \times 303}{288 \times 2} \\ \\ \dashrightarrow \tt \: V = \dfrac{4.1\times 2.4 \times 303}{288} \\ \\ \dashrightarrow \tt \: V = \dfrac{4.1 \times 2.4 \times 101}{96} \\ \\ \dashrightarrow \tt \: V = \dfrac{4.1 \times 2.4 \times 37}{32} \\ \\ \dashrightarrow \tt \: V = \dfrac{4.1 \times 0.3 \times 37}{4} \\ \\ \dashrightarrow \tt \: V =4.1 \times 0.3 \times 9.25 \\ \\ \dashrightarrow \boxed{ \boxed{\tt \: V_{final} =11.37 \: {cm}^{3} }}$

Answer 2

Given that , A small bubble rise from a bottom of the sea where the temperature is 15°C and pressure of 8.2 atm reaches the surface of sea where the temperature is 30°C and pressure of 2 atm & The initial volume is 2.4 cm³ .

Exigency To Find : The Final Volume ?

⠀⠀⠀⠀⠀━━━━━━━━━━━━━━━━━━━⠀

⠀⠀⠀Given that ,

⠀⠀⠀⠀⠀▪︎⠀Initial Volume ( V₁ ) is 2.4 cm³ .

⠀⠀⠀⠀⠀▪︎⠀Initial Temperature ( T₁ ) is 15°C [ Temperature at bottom of sea ]

⠀⠀⠀⠀⠀▪︎⠀Initial Pressure ( P₁ ) is 8.2 atm . [ Pressure at bottom of sea ]

⠀⠀⠀⠀⠀▪︎⠀Final Temperature ( T₂ ) is 30°C [ Temperatures at surface of sea ]

⠀⠀⠀⠀⠀▪︎⠀Final Pressure ( P₂ ) is 2 atm . [ Pressure at surface of sea ]

⠀⠀⠀⠀⠀⠀⠀⠀⠀¤ Converting Temperature unit from Celsius ( C ) to Kelvin ( K ) :

$\qquad \dashrightarrow \sf Initial \:Temperature \:\:( \:T_1 \:) \:=\: 15^0 C \:\\\\\qquad \dashrightarrow \sf Initial \:Temperature \:\:( \:T_1 \:) \:=\:\Big\{ 15 + 273 \:\Big\} \:K\: \qquad \:\bigg\lgroup \sf{ Kelvin\:(K)\:=\:Celsius\:(C) \: + 273 \: }\bigg\rgroup\:\\\\\qquad \dashrightarrow \sf Initial \:Temperature \:\:( \:T_1 \:) \:=\:\Big\{ 288 \:\Big\} \:K\: \:\\\\\qquad \dashrightarrow \sf Initial \:Temperature \:\:( \:T_1 \:) \:=\: 288 \: \:K\: \:\\\\\qquad \dashrightarrow \underline {\boxed{\pmb{\frak{\purple { \:\:Initial \:Temperature \:\:( \:T_1 \:) \:=\: 288 \: \:K\: \:}}}}}\:\:\bigstar$

⠀⠀⠀⠀⠀⠀⠀⠀⠀AND ,

$\qquad \dashrightarrow \sf Final \:Temperature \:\:( \:T_2 \:) \:=\: 30^0 C \:\\\\\qquad \dashrightarrow \sf Final \:Temperature \:\:( \:T_2 \:) \:=\:\Big\{ 30 + 273 \:\Big\} \:K\: \qquad \:\bigg\lgroup \sf{ Kelvin\:(K)\:=\:Celsius\:(C) \: + 273 \: }\bigg\rgroup\:\\\\\qquad \dashrightarrow \sf Final \:Temperature \:\:( \:T_2 \:) \:=\:\Big\{ 303 \:\Big\} \:K\: \:\\\\\qquad \dashrightarrow \sf Final \:Temperature \:\:( \:T_2 \:) \:=\: 303 \: \:K\: \:\\\\\qquad \dashrightarrow \underline {\boxed{\pmb{\frak{\purple { \:\:Final \:Temperature \:\:( \:T_2 \:) \:=\: 303 \: \:K\: \:}}}}}\:\:\bigstar$

⠀⠀⠀⠀⠀━━━━━━━━━━━━━━━━━━━⠀

❍ Let's Consider Final Volume be V₂ .

As , We know that,

⠀⠀⠀⠀⠀▪︎⠀IDEAL GAS EQUATION :

$\qquad \dag\:\:\bigg\lgroup \pmb{\bf \:\:PV\:=\:nRT \: \:\:}\bigg\rgroup$

⠀⠀⠀⠀⠀⠀⠀⠀⠀Here , P is the Pressure , T is the Temperature in Kelvin , V is the Golume , n is the number of molecules & R is the Universal gas Constant.

$\qquad \dashrightarrow \sf \:\:PV\:\:nRT \: \\\\\qquad \dashrightarrow \sf \:\:\dfrac{PV}{T}\:=\:nR \:$

• nR remains constant .

$\qquad \dashrightarrow \sf \:\:\dfrac{PV}{T}\:=\:nR \: \\\\\qquad \dashrightarrow \sf \:\:\dfrac{PV}{T}\: \:$

Therefore,

$\qquad \dashrightarrow \sf \:\:\Big\{ \dfrac{PV}{T}\:\Big\}\:\:_1 \:=\:\:\:\Big\{ \dfrac{PV}{T}\:\Big\}\:\:_2 \: \:$

$\qquad \dashrightarrow \sf \:\:\Big\{ \dfrac{P_1 \times V_1 }{T_1 }\:\Big\}\:\: \:=\:\:\:\Big\{ \dfrac{P_2 \times V_2 }{T_2 }\:\Big\}\:\: \: \:$

⠀⠀Where ,

⠀⠀⠀⠀⠀▪︎⠀Initial Volume ( V₁ ) is 2.4 cm³ .

⠀⠀⠀⠀⠀▪︎⠀Initial Temperature ( T₁ ) is 15°C .

⠀⠀⠀⠀⠀▪︎⠀Initial Pressure ( P₁ ) is 8.2 atm .

⠀⠀⠀⠀⠀▪︎⠀Final Temperature ( T₂ ) is 30°C .

⠀⠀⠀⠀⠀▪︎⠀Final Pressure ( P₂ ) is 2 atm .

⠀⠀⠀⠀⠀▪︎⠀Final Volume ( V₂ ) is ??

⠀⠀⠀⠀⠀⠀$\\\qquad \underline {\boldsymbol{\star\:Now \: By \: Substituting \: the \: known \: Values \::}}$

$\qquad \dashrightarrow \sf \:\:\Big\{ \dfrac{P_1 \times V_1 }{T_1 }\:\Big\}\:\: \:=\:\:\:\Big\{ \dfrac{P_2 \times V_2 }{T_2 }\:\Big\}\:\: \: \: \\\\\qquad \dashrightarrow \sf \:\:\Big\{ \dfrac{ 8.2 \times 2.4 }{288 }\:\Big\}\:\: \:=\:\:\:\Big\{ \dfrac{ 2 \times V_2 }{ 303 }\:\Big\}\:\: \: \: \\\\\qquad \dashrightarrow \sf \:\:V_2 \:=\: \dfrac{ 8.2 \times 2.4 \: \times 303 \:}{288\times 2 \:} \: \: \\\\\qquad \dashrightarrow \sf \:\:V_2 \:=\: \dfrac{ 4.1 \times 2.4 \: \times 303 \:}{288 \:} \: \: \\\\\qquad \dashrightarrow \sf \:\:V_2 \:=\: \dfrac{ 4.1 \times 2.4 \: \times 101\:}{96 \:} \: \: \\\\\qquad \dashrightarrow \sf \:\:V_2 \:=\: \dfrac{ 4.1 \times 2.4 \: \times 37\:}{32 \:} \: \: \\\\\qquad \dashrightarrow \sf \:\:V_2 \:=\: \dfrac{ 4.1 \times 0.3 \: \times 37\:}{4 \:} \: \: \\\\\qquad \dashrightarrow \sf \:\:V_2 \:=\: 4.1 \times 0.3 \: \times 9.25 \: \: \\\\\qquad \dashrightarrow \sf \:\:V_2 \:=\: 11.37 \: \: \\\\\qquad \dashrightarrow \underline {\boxed{\pmb{\frak{\purple { \:\:Final \:Volume \:\:( \:V_2 \:) \:=\: 11.37 \: \:cm^3\: \:}}}}}\:\:\bigstar$

$\qquad \therefore \underline {\sf Hence, \:The \: Final \:Volume \:is \:\pmb{\bf 11.37 \:cm^3\:}\:\:}$