Suppose the Gas is acting ideally, Then, According to Ideal Gas Equation;
P₁ V₁ = P₂ V₂ ----- (1)
Data Given:
P₁ = 885 torr
V₁ = 125 cm³
P₂ = 225 torr
V₂ = ?
Puting values in eq, 1;
V₂ = P₁ V₁ / P₂
V₂ = (885 torr × 125 cm³) ÷ 225 torr
V₂ = 491.66 cm³
The heated air temperature from inside the balloon more likely rises since it can't escape its trapped resulting in different density from the cooler air, the cooler air from the outside is a bit denser than the heated.
Answer:
Increase the amplitude by blowing with more force.
Explanation: I answered this.
A sour-tasting material (usually in a solution) that dissolves metals and other materials. Technically, a material that produces positive ions in solution. An acid<span> is the opposite of a base and has a pH of 0 to 7.</span>
Answer:
1. n = 0.174mol
2. T= 26.8K
3. P = 1.02atm
4. V = 126.88L
Explanation:
1. P= 2.61atm
V = 1.69L
T = 36.1 °C = 36.1 + 273= 309.1K
R = 0.082atm.L/mol /K
n =?
n = PV / RT = (2.61x1.69)/(0.082x309.1)
n = 0.174mol
2. P = 302 kPa = 302000Pa
101325Pa = 1atm
302000Pa = 302000/101325 = 2.98atm
V = 2382 mL = 2.382L
T =?
n = 3.23 mol
R = 0.082atm.L/mol /K
T= PV /nR = (2.98x2.382)/(3.23x0.082) = 26.8K
3. P =?
V = 0.0250 m³ = 25L
T = 288K
n = 1.08mol
R = 0.082atm.L/mol /K
P = nRT/V = (1.08x0.082x288)/25 = 1.02atm
4. P = 782 torr
760Torr = 1 atm
782 torr = 782/760 = 1.03atm
V =?
T = 303K
n = 5.26 mol
R = 0.082atm.L/mol /K
V = nRT/P
V = (5.26x0.082x303)/1.03 = 126.88L
