A gas storage tank has a volume of 3.5*10^5 m^3 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 to -10 degrees celcius and the pressure drops to 95 kPa?
Note the correct spelling of celsius.
Use (P1V1)/T1 = (P2V2)/T2
Don't forget to change T to Kelvin.
To solve this problem, 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
T is the temperature in Kelvin
First, let's convert the original temperature and pressure to Kelvin and kPa:
Temperature (T1) = 27 degrees Celsius = 27 + 273.15 = 300.15 K
Pressure (P1) = 101 kPa
Using the ideal gas law, we can calculate the number of moles (n1) of gas in the tank at the initial conditions:
n1 = (P1 * V1) / (R * T1)
Next, let's convert the new temperature and pressure to Kelvin and kPa:
Temperature (T2) = -10 degrees Celsius = -10 + 273.15 = 263.15 K
Pressure (P2) = 95 kPa
Using the ideal gas law, we can calculate the new volume (V2) of the tank at the new conditions:
V2 = (n1 * R * T2) / P2
Substituting the values we calculated:
V2 = ((P1 * V1) / (R * T1)) * (R * T2) / P2
Now let's solve the equation step-by-step:
Step 1: Convert the values given in the problem to SI units
T1 = 300.15 K
P1 = 101 kPa = 101000 Pa
T2 = 263.15 K
P2 = 95 kPa = 95000 Pa
Step 2: Use the ideal gas law equation to calculate n1
n1 = (P1 * V1) / (R * T1)
Step 3: Substitute the known values into the equation and solve for n1
n1 = (101000 Pa * 3.5 * 10^5 m^3) / (8.314 J/(mol·K) * 300.15 K)
Step 4: Calculate n1
n1 = 14152.28 mol
Step 5: Use the ideal gas law equation to calculate V2
V2 = (n1 * R * T2) / P2
Step 6: Substitute the known values into the equation and solve for V2
V2 = (14152.28 mol * 8.314 J/(mol·K) * 263.15 K) / 95000 Pa
Step 7: Calculate V2
V2 = 312.24 m^3
Therefore, the new volume of the tank is 312.24 cubic meters when the temperature drops to -10 degrees Celsius and the pressure drops to 95 kPa.
To find the new volume of the tank, we can use the ideal gas law, which states:
PV = nRT
Where:
P is the pressure of the gas
V is the volume of the gas
n is the number of moles of gas
R is the ideal gas constant (8.314 J/(mol*K))
T is the temperature of the gas in Kelvin
To solve this problem, we need to convert the temperatures from Celsius to Kelvin. The conversion formula is:
T(K) = T(°C) + 273.15
Given:
Initial volume (V1) = 3.5 * 10^5 m^3
Initial temperature (T1) = 27 °C
Initial pressure (P1) = 101 kPa
New temperature (T2) = -10 °C
New pressure (P2) = 95 kPa
Converting the initial temperature to Kelvin:
T1(K) = 27 + 273.15 = 300.15 K
Converting the new temperature to Kelvin:
T2(K) = -10 + 273.15 = 263.15 K
Now, we need to compare the initial and final states of the gas, keeping the number of moles and the ideal gas constant the same. Since n and R do not change in this problem, we can rearrange the formula to solve for the new volume (V2):
V2 = (P2 * V1 * T1) / (P1 * T2)
Substituting the given values into the formula:
V2 = (95 kPa * 3.5 * 10^5 m^3 * 300.15 K) / (101 kPa * 263.15 K)
Now, let's calculate the new volume:
V2 = (95 * 300.15 * 3.5 * 10^5) / (101 * 263.15) = 3.381 * 10^5 m^3
Therefore, the new volume of the tank is approximately 3.381 * 10^5 m^3 when the temperature drops to -10 degrees Celsius and the pressure drops to 95 kPa.