A gas storage tank has a volume of 350,000 when the temperature is 27 degrees celcius and the pressure is 101 kPa. what is the new volume of the tank if the temperature drops -10 degrees celcius and the pressure drops to 95 kPa? please show work.
Moles didn't change
P1V1/T1=P2V2/T2
temps in kelvins.
solvefor V2.
To find the new volume of the gas storage tank, we need to use the ideal gas law, which is given by the formula:
PV = nRT
Where:
P = Pressure
V = Volume
n = moles of gas
R = ideal gas constant
T = Temperature in Kelvin
First, let's convert the temperatures to Kelvin. The formula to convert from Celsius to Kelvin is:
T(K) = T(°C) + 273.15
Given:
Initial temperature (T₁) = 27 °C
Final temperature (T₂) = 27 - 10 = 17 °C
Therefore:
Initial temperature (T₁) in Kelvin = 27 + 273.15 = 300.15 K
Final temperature (T₂) in Kelvin = 17 + 273.15 = 290.15 K
Now, let's calculate the new volume of the tank.
Since the number of moles of gas and the gas constant are constant, we can rewrite the equation as:
P₁V₁ / T₁ = P₂V₂ / T₂
Where:
P₁ = Initial pressure = 101 kPa
V₁ = Initial volume = 350,000 L
P₂ = Final pressure = 95 kPa
V₂ = Final volume (to be determined)
Substituting the given values:
(101 kPa)(350,000 L) / (300.15 K) = (95 kPa)(V₂) / (290.15 K)
Now we can solve for V₂:
V₂ = [(101 kPa)(350,000 L)(290.15 K)] / [(300.15 K)(95 kPa)]
V₂ ≈ (9.4365 × 10^9 kPa * L) / (2.8558 × 10^4 kPa)
V₂ ≈ 330,285.3 L
Therefore, the new volume of the tank, when the temperature drops to -10 °C and the pressure drops to 95 kPa, is approximately 330,285.3 L.