Vol.250 before its to much pressure
Answer: 94.13 L
Explanation: In STP in an ideal gas there is a standard value for both temperature and pressure. At STP,pressure is equal to 1atm and the temperature at 0°C is equal to 273.15K. This problem is an ideal gas so we use PV=nRT where R is a constant R= 0.08205 L.atm/mol.K.
To find volume, derive the equation, it becomes V=nRT/P. Substitute the values. V= 4.20 mol( 0.08205L.atm/mol.K)(273.15K) / 1 atm = 94.13 L. The mole units, atm and K will be cancelled out and L will be the remaining unit which is for volume.
Answer:
2.82 L
T₁ = 303 K
T₂ = 263 K
The final volume is smaller.
Explanation:
Step 1: Given data
- Initial temperature (T₁): 30 °C
- Initial volume (V₁): 3.25 L
- Final temperature (T₂): -10 °C
Step 2: Convert the temperatures to Kelvin
We will use the following expression.
K = °C + 273.15
T₁: K = 30°C + 273.15 = 303 K
T₂: K = -10°C + 273.15 = 263 K
Step 3: Calculate the final volume of the balloon
Assuming constant pressure and ideal behavior, we can calculate the final volume using Charles' law. Since the temperature is smaller, the volume must be smaller as well.
V₁/T₁ = V₂/T₂
V₂ = V₁ × T₂/T₁
V₂ = 3.25 L × 263 K/303 K = 2.82 L
In hot water the molecules move faster versus In cold water they move slower (hope that helps)
