Question

A sample of helium occupies a volume of 160cm3 at 100 KPa and 25°c. what volume will it occupy if the pressure is adjusted to 80 KPa and the temperature remains unchanged?​

Answer 1

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Answer 2

Boyle's Law-

`\:\:\:\:\:\:\:\:\:\:\:\star\:\sf \underline{ P_1 \: V_1=P_2 \: V_2}\\`

(Pressure is inversely proportional to the volume)

Where-

• 
  `\sf V_1` = Initial volume
• 
  `\sf V_2` = Final volume
• 
  `\sf P_1` = Initial pressure
• 
  `\sf P_2` = Final pressure

As per question, we are given that -

• 
  `\sf V_1` = 160 cm³
• 
  `\sf P_1` = 100KPa
• 
  `\sf P_2` = 80KPa

Now that we have all the required values and we are asked to find out that volume which will be occupied if the pressure is adjusted to 80 KPa and the temperature remains unchanged. For that we can put the values and solve for the final volume of helium-

`\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\star\:\sf \underline{ P_1 \: V_1=P_2 \: V_2}`

`\:\:\:\: \:\:\:\:\:\:\longrightarrow \sf 100 * 160 = 80 * V_2\\`

`\:\:\:\:\:\:\:\:\:\:\longrightarrow \sf V_2 = (100 * 160)/(80)\\`

`\:\:\:\:\:\:\:\:\:\:\longrightarrow \sf V_2 =100* \cancel{( 160)/(80)}\\`

`\:\:\:\: \:\:\:\:\:\:\longrightarrow \sf V_2 = 100 * 2\\`

`\:\:\:\:\:\:\:\:\:\:\longrightarrow \sf \underline{V_2 = 200 \:cm^3 }\\`

• Therefore, 200 cm³ will be occupied if the pressure is adjusted to 80 KPa and the temperature remains unchanged.