Question

What volume did a helium-filled balloon have at 22.5 c and 1.95 atm if it’s new volume was 56.4 mL at 3.69 atm and 11.9c

Answer 1

111 ml

Step-by-step explanation:

Use the general rule for gases:

P1 V1 / T1 = P2 V2 / T2 Looking for V1 T must be in Kelvin

re-arrange to :

V1 = P2 V 2 * T1 / (T2 * P1) <==== now sub in the values

V1 = 3.69 * 56.4 * (22.5 + 273.15) / [(11.9 + 273.15) * 1.95]

V1 = 110.6949 ml

for correctness, the answer should only have THREE significant digits as all of the factors have three SD

V1 = 111 ml

Answer 2

This is an exercise in the general or combined gas law.

To start solving this exercise, we obtain the following data:

Data:

• T₁ = 22.5 °C + 273 = 295.5 K
• P₁ = 1.95 atm
• V₁ = ¿?
• P₂ = 3.69 atm
• T₂ = 11.9 °C + 273 = 284.9 k
• V₂= 56.4 ml

We use the following formula:

P₁V₁T₂ = P₂V₂T₁ ⇒ General formula

Where

• P₁ = Initial pressure
• V₁ = Initial volume
• T₂ = Initial temperature
• P₂ = Final pressure
• V₂ = final volume
• T₁ = Initial temperature

We clear the formula for the initial volume:

`\boldsymbol{\sf{V_(1)=(P_(2)V_(2)T_(1))/(P_(1)T_(2)) } }`

We substitute our data into the formula to solve:

`\boldsymbol{\sf{V_(1)=\frac{(3.69 \\ot{atm})(56.4 \ ml)(295.5 \\ot{k})}{(1.95 \\ot{atm})(284.9\\ot{k})} }}`

`\boldsymbol{\sf{V_(1)=(61498.278)/(555.555) \ lm }}`

`\boxed{\boldsymbol{\sf{V_(1)=110.697 \ lm }}}`

The helium-filled balloon has a volume of 110.697 ml.