A large balloon has a volume of 2.50 L at 0 C. At what temperature 0 C will the volume of the gas be 3.60 L Assume constant temperature
To solve this problem, we can use Charles' Law, which states that at constant pressure, the volume of a gas is directly proportional to its temperature measured on the Kelvin scale.
First, we need to convert the initial temperature of 0 °C to Kelvin. To do this, we add 273.15 to the Celsius temperature:
Initial temperature (T1) = 0 °C + 273.15 = 273.15 K
Initial volume (V1) = 2.50 L
Next, let's define the variables:
Final temperature (T2) = unknown
Final volume (V2) = 3.60 L
Now, we can set up the equation using Charles' Law:
V1 / T1 = V2 / T2
Plugging in the given values in accordance with the equation:
2.50 L / 273.15 K = 3.60 L / T2
To isolate T2, we can cross-multiply the equation:
2.50 L * T2 = 3.60 L * 273.15 K
Now, divide both sides of the equation by 2.50 L to solve for T2:
T2 = (3.60 L * 273.15 K) / 2.50 L
Calculating the equation:
T2 = 393.9 K
Finally, to convert the temperature back to Celsius, subtract 273.15 from the Kelvin temperature:
T2 = 393.9 K - 273.15 = 120.75 °C
Therefore, the volume of the gas will be 3.60 L at a temperature of 120.75 °C.