Answer:
45.0 L is the volume of gas will the balloon contain at 1.35 atm and 253 K.
Explanation:
Using Ideal gas equation for same mole of gas as
Given ,
V₁ = 25.0 L
V₂ = ?
P₁ = 2575 mm Hg
Also, P (atm) = P (mm Hg) / 760
P₁ = 2575 / 760 atm = 3.39 atm
P₂ = 1.35 atm
T₁ = 353 K
T₂ = 253 K
Using above equation as:
Solving for V₂ , we get:
<u>V₂ = 45.0 L</u>
45.0 L is the volume of gas will the balloon contain at 1.35 atm and 253 K.
The answer for the following problem has been mentioned below.
Explanation:
Given:
volume of oxygen (V) = 4.50 L
Temperature (T) = 425 K
pressure of oxygen (P) = 2.50 atm
Gram molecular mass of oxygen (M) = 16.0 grams
To calculate:
mass of water (m)
We know;
According to the ideal gas equation;
P × V = n × R × T
As we know;
no of moles =
m represents the mass of oxygen (m)
M represents the Gram molecular mass (M)
According to above mentioned equation;
P × V = n × R × T
P represents the pressure of the oxygen
V represents the volume of the oxygen
n represents the no of moles of the oxygen
R represents the universal gas constant
where,
the value of R is 0.0821 L atm/K moles
Substituting the values in the above equation;
2.50 × 4.50 = × 0.0821 × 425
11.25 = × 34.8925
180 = m × 34.8925
m =
m = 5.158 grams
Therefore the mass of the of oxygen is 5.158 grams
Now;
As we know;
where;
represents the mass of the oxygen
represents the gram molecular mass of the oxygen
represents the mass of the water
represents the gram molecular mass of water
From the above given formula,
=
where;
Gram molecular weight of water = 18.0 u
= 5.802 grams
<em><u>Therefore the mass of the water is 5.802 grams.</u></em>