Question

If 600 L of oxygen was collected when the temperature is 27oC, what will the volume be at 23oC?

Answer 1

Given :

• Initial volume when oxygen was collected ( V₁ ) = 600 L
• Initial temperature when oxygen was collected ( T₁ ) = 27° C
• Final temperature of oxygen collected ( T₂ ) = 23°C

To find :

Final volume ( V₂ ) of oxygen collected

Solution :

Note - As we are not given any information about pressure, we will assume that pressure remains same during whole process.

According to Charles's law , under constant pressure volume of gas is directly proportional to its temperature in Kelvin scale.

➝ V ∝ T

$\sf \implies \:\dfrac{V_1}{V_2} \: = \: \dfrac{T_1}{T_2}$

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Now. first of all we need to convert temperature into Kelvin scale.

• T₁ = 27° C

➝ T₁ = 27 + 273 K

➝ T₁ = 300 K

• T₂ = 23°C

➝ T₂ = 23° + 273

➝ T₂ = 296 K

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$\sf \implies \:\dfrac{600\:L}{V_2} \: = \: \dfrac{300\:K}{296\:K}$

$\sf \implies \:V_2\: = \: \dfrac{296\:K}{300\:K} \times 600 \: L$

$\sf \implies \:V_2\: = \: 296 \: \times \: \dfrac{600\:}{300\:} \: L$

$\sf \implies \:V_2\: = \: 296 \: \times 2 \: L$

$\sf \implies \:V_2\: = \: \bold{ 592 \: L}$

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ANSWER : 592 L