A balloon with a volume of 950. mL is cooled from 345 K to 298 K. The internal pressure does not change and the balloon does not leak. What is the volume of the cooled gas?
I think I got the Right answer but I am a Little Confused and online school is Making this Confusing for Me
would it be 820.58 mL?
To find the volume of the cooled gas, we can use the ideal gas law equation: PV = nRT, where P is the pressure, V is the volume, n is the number of moles of gas, R is the ideal gas constant, and T is the temperature in Kelvin.
In this case, we know that the internal pressure does not change, so we can consider the pressure as constant. The number of moles of gas can also be considered constant since the balloon does not leak.
We can rewrite the ideal gas law equation as V1/T1 = V2/T2, where V1 and T1 are the initial volume and temperature, and V2 and T2 are the final volume and temperature.
Given:
V1 = 950 mL = 950 cm3 (since 1 mL = 1 cm3)
T1 = 345 K
T2 = 298 K
Now we can plug in the values into the equation and solve for V2:
V1/T1 = V2/T2
950 cm3 / 345 K = V2 / 298 K
To solve for V2, we can cross-multiply and rearrange the equation:
V2 = 950 cm3 * 298 K / 345 K
Performing the calculation:
V2 = 820.87 cm3
Therefore, the volume of the cooled gas is approximately 820.87 cm3.
(V1/T1) = (V2/T2)
(950/345) = (V2/298)
Solve for V2