What pressure is required to reduce 70 mL of
a gas at standard conditions to 16 mL at a
temperature of 22◦C? Answer in atm.
ASAP, its due tomorrow
Boyles Law
P₁∙V₁ = P₂∙V₂
V₁ = 70ml
P₁ =760mmHg
V₂ = 16ml
P₂ = ?
T₁ = T₂ = 22⁰ = 295K = Constant
Substitute given values and solve for P₂.
P₂ = 3325mmHg = 4.375Atm or, 1 sig.figs. (V₁ = 70ml) => 3000mmHg = 4Atm
To answer this question, we can use the combined gas law, which relates the initial and final conditions of a gas sample. The combined gas law is expressed as follows:
(P₁ × V₁) / (T₁) = (P₂ × V₂) / (T₂)
Where:
P₁ = initial pressure (atm)
V₁ = initial volume (mL)
T₁ = initial temperature (K)
P₂ = final pressure (atm)
V₂ = final volume (mL)
T₂ = final temperature (K)
In this case, we are given the following information:
V₁ = 70 mL
V₂ = 16 mL
T₁ = 273 + 22 = 295 K (since the temperature should be in Kelvin)
We need to find P₂, the final pressure in atm.
Now, let's plug in the values into the equation:
(P₁ × V₁) / (T₁) = (P₂ × V₂) / (T₂)
P₂ = [(P₁ × V₁) / (V₂)] × [(T₂) / (T₁)]
Since the gas is at standard conditions, we know that the initial pressure is 1 atm. Plugging in the values:
P₂ = [(1 atm × 70 mL) / (16 mL)] × [(295 K) / (273 K)]
Simplifying the equation:
P₂ = (70/16) × (295/273) atm
Calculating the expression:
P₂ ≈ 2.016 atm
Therefore, the pressure required to reduce 70 mL of gas at standard conditions to 16 mL at a temperature of 22◦C is approximately 2.016 atm.