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

Answer:

110.69 ml

Explanation:

Not sure why my answer was removed:

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 = 110.69 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:

$\boldsymbol{\sf{V_{1}=\dfrac{P_{2}V_{2}T_{1}}{P_{1}T_{2}} } }$

We substitute our data into the formula to solve:

$\boldsymbol{\sf{V_{1}=\dfrac{(3.69 \not{atm})(56.4 \ ml)(295.5 \not{k})}{(1.95 \not{atm})(284.9\not{k})} }}$

$\boldsymbol{\sf{V_{1}=\dfrac{61498.278}{555.555} \ lm }}$

$\boxed{\boldsymbol{\sf{V_{1}=110.697 \ lm }}}$

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