Question

$\huge\sf\fbox\pink{QUESTION}$To what temperature must a gas at 127 °C be cooled, so that its volume is reduced to 1/5 of the initial volume?

Note :
• No spam
• Correct answer required
• Step-by-explanation
• Correct answer will be mark as brainliest.​

Answer 1

It can be solved by using the relation of

Charle's Law. Charle's Law explains that:

For a fixed mass of gas at constant

pressure (P) ,the volume is directly

proportional to the kelvin temperature

(T). T2 = ? So the temperature should be

reduced to 100 K to get the 1/4 of initial

volume of gas.

Answer 2

$\bold{\huge{\color{aqua}{solution:-}}}$

$\bold{\small\blue{Let\:us\:take\:initial\: volume\:(V_1)=x}}$

$\bold{\small\blue{Initial\: temperature\:(T_1)=127°C=400\:K}}$

$\bold{\blue{\small Final\: volume\:(V_2)=\frac{x}{5}}}$

$\bold{\purple{We\:know \:that}}$

$\bold{\blue{\small Final\: temperature\:(T_2)=\orange{\frac{V_2}{V_1}×T_1}}}$

$\bold{\purple{On\: substituting\:value }}$

$\bold{\Rightarrow \small T_2=\frac{\frac{x}{5}}{x}×400}$

$\bold{\Rightarrow \small T_2=\frac{x×400}{5×x}}$

$\bold{\Rightarrow \small T_2=\frac{\red{\cancel{\blue{x}}}×\red{\cancel{\green{400}}}^{\:\blue{80}}}{\red{\cancel{\green{\:\:5}}}×\red{\cancel{\blue{x}}}}}$

$\bold{\Rightarrow\small T_2=80 \:K}$

$\bold{\Rightarrow T_2=\small (80-273)°C}$

$\bold{\Rightarrow \green{\boxed{\red{T_2}=\pink{-193°C}}}}$