Question

500 ml of a certain gas at 30°C was cooled to -20°C by keeping the pressure
constant. Find the contraction in volume.​

Answer 1

Answer:-

Contraction in volume is 82.5 mL

Explanation:-

We have :-

→ Initial volume (V₁) = 500 mL

→ Initial temperature (T₁) = 30°C

→ Final temperature (T₂) = -20°C

_______________________________

Firstly, let's convert initial and final temperature from °C to K.

Initial:-

⇒ 0° C = 273 K

⇒ 30°C = 273 + 30

⇒ 303 K

Final:-

⇒ -20°C = 273 - 20

⇒ 253 K

________________________________

According to Charle's Law, we know that :-

V₁/T₁ = V₂/T₂ [Pressure is constant]

Substituting values, we get :-

⇒ 500/303 = V₂/253

⇒ V₂ = 500 × 253/303

⇒ V₂ = 126500/303

⇒ V₂ = 417.5 mL

Now, contraction in volume :-

= V₁ - V₂

= (500 - 417.5) mL

= 82.5 mL

Answer 2

${\large{\pmb{\sf{\bigstar \: {\underline{Understanding \: the \: Question...}}}}}}$

★ This question says that we have to find out the contraction in volume for 500 millilitres of a certain gas at 30 degrees celcius and it was cooled to -20 degrees celcius. Let's solve this!

${\large{\pmb{\sf{\bigstar \: {\underline{Given \; that...}}}}}}$

★ Gas = 500 millilitres

★ Temperature of gas = 30°C

★ Gas was cooled to -20°C from 30°C

${\large{\pmb{\sf{\bigstar \: {\underline{To \; find...}}}}}}$

★ Contraction in volume

${\large{\pmb{\sf{\bigstar \: {\underline{Solution...}}}}}}$

★ Contraction in volume = 82.51 ml

${\large{\pmb{\sf{\bigstar \: {\underline{Using \; concepts...}}}}}}$

★ Formula to find out the new volume =

${\small{\underline{\boxed{\sf{\dfrac{V_1}{T_1} \: = \dfrac{V_2}{T_2}}}}}}$

★ Formula to change Celcius to Kelvin =

${\small{\underline{\boxed{\sf{0 \degree Celcius \: = 273 \: Kelvin}}}}}$

${\large{\pmb{\sf{\bigstar \: {\underline{Where...}}}}}}$

★ V₁ dentoes volume no. one

★ T₁ dentoes temperature no. one

★ V₂ dentoes volume no. two

★ T₂ dentoes temperature no. two

${\large{\pmb{\sf{\bigstar \: {\underline{Here...}}}}}}$

★ V₁ is 500 ml

★ T₁ is 30°C (303 K)

★ V₂ is ?

★ T₂ is -20°C (253 K)

${\large{\pmb{\sf{\bigstar \: {\underline{Full \; Solution...}}}}}}$

~ Firstly we have to convert Celcius into Kevin's. Let us convert them!

${\small{\underline{\boxed{\sf{0 \degree Celcius \: = 273 \: Kelvin}}}}} \\ \\ :\implies \sf 0 \degree Celcius \: = 273 \: Kelvin \\ \\ :\implies \bf Changing \: T_1 \: firstly \\ \\ :\implies \sf 273 + 30 \\ \\ :\implies \sf 303 \: Kelvin \\ \\ :\implies \bf Changing \: T_2 \: now \\ \\ :\implies \sf 273 - 20 \\ \\ :\implies \sf 253 \: Kevin$

~ Now we have to use this law's rule that is mentioned below!

${\small{\underline{\boxed{\sf{\dfrac{V_1}{T_1} \: = \dfrac{V_2}{T_2}}}}}} \\ \\ :\implies \sf \dfrac{V_1}{T_1} \: = \dfrac{V_2}{T_2} \\ \\ :\implies \sf \dfrac{500}{303} \: = \dfrac{V_2}{253} \\ \\ :\implies \sf 1.65 \: = \dfrac{V_2}{253} \\ \\ :\implies \sf 1.65 \times 253 \: = V_2 \\ \\ :\implies \sf 417.49 \: = V_2\\ \\ :\implies \sf \: V_2 = 417.49 \: ml$

~ Now at last let us find the contraction in volume!

${\small{\underline{\boxed{\sf{V_1 - V_2}}}}} \\ \\ :\implies \sf V_1 - V_2 \\ \\ :\implies \sf 500 - 417.49 \\ \\ :\implies \sf Contraction \: = 82.51 \: ml$

${\large{\pmb{\sf{\bigstar \: {\underline{Additional \; Knowledge...}}}}}}$

$\begin{gathered}\boxed{\begin {array}{c|c|c|c}\sf Temperature \:Scale &\sf Ice\:point &\sf Steam \;point &\sf No\;of\:division\:on\:Fundamental\;interval \\ \frac {\qquad\qquad\qquad}{} &\frac{\qquad\qquad\qquad}{} & \frac {\qquad\qquad\qquad}{} & \frac {\qquad\qquad\qquad\qquad\qquad\qquad}{} \\ \sf Celsius\;scale & \sf0^\circ C &\sf100^\circ C &\sf 100\\&&&\\ \sf Fahrenheit\:scale & \sf32^\circ F &\sf 212^\circ F &\sf180 \\&&&\\ \sf Kelvin\;scale &\sf273K &\sf373K & \sf100\\&&&\\ \sf Reaumer \:scale &\sf0^\circ R &\sf80^\circ R &\sf80\end {array}}\end{gathered}$