a 2.00 L ballon at 25 degrees celcius and 745mmHg floats on an altitude where the temperature is 10 degrees celcius and the air pressure is 700mmHg. what is the new volume othe balloon
Use the combined gas law:
P1V1/T1=P2V2/T2
Where
P1=745mmHg
V1=2.00L
T1=273K+25=298.15K
P2=700mmHg
V2=?
T2=273.15+10=283.15K
Solve for V2:
V2=(P1/P2)*(T2/T1)*V1
Answer should contain three significant figures.
1.06L
Convert pressures from mmHg to atm then use combined gas law to solve for the new volume. You get 2.06 Liters
To find the new volume of the balloon, we can use the combined gas law equation, which combines Boyle's law and Charles's law:
(P₁V₁) / (T₁) = (P₂V₂) / (T₂)
Where:
P₁ and P₂ are the initial and final pressures (in mmHg).
V₁ and V₂ are the initial and final volumes (in liters).
T₁ and T₂ are the initial and final temperatures (in Kelvin).
First, let's convert the temperatures from Celsius to Kelvin:
Initial temperature (T₁) = 25°C + 273.15 = 298.15 K
Final temperature (T₂) = 10°C + 273.15 = 283.15 K
Next, we convert the pressures to the same units (mmHg):
Initial pressure (P₁) = 745 mmHg
Final pressure (P₂) = 700 mmHg
Now, we can substitute these values into the combined gas law equation and solve for V₂ (the final volume of the balloon):
(P₁V₁) / T₁ = (P₂V₂) / T₂
V₂ = (P₁V₁ * T₂) / (P₂ * T₁)
Plugging in the values we have:
V₂ = (745 mmHg * 2.00 L * 283.15 K) / (700 mmHg * 298.15 K)
Calculating this equation gives us the new volume of the balloon.