a balloonist is preparing a trip in a helium balloon. the trip begins in early morning at a temp of 15 degrees Celcius. By mid afternoon, the temp has increased to 30 degrees C. assuming the pressure remains constant at 1.00 atm, for each mole of helium, calculate:
1. the initial and final volume
2. the change in internal energy
3. the work done by helium
4. change in enthalpy
To answer these questions, we need to use the Ideal Gas Law, which relates the pressure, volume, temperature, and the number of moles of a gas. The Ideal Gas Law can be written as:
PV = nRT
Where:
P = pressure (in atm)
V = volume (in liters)
n = number of moles
R = gas constant (0.0821 L·atm/mol·K)
T = temperature (in Kelvin)
Let's calculate each of the quantities step-by-step:
1. The Initial and Final Volume:
We are given the initial and final temperatures: 15 degrees Celsius and 30 degrees Celsius, respectively. We need to convert these temperatures to Kelvin by adding 273.15 to each.
Initial temperature (T1) = 15 + 273.15 = 288.15 K
Final temperature (T2) = 30 + 273.15 = 303.15 K
Since the pressure (P) is constant at 1 atm, we can rearrange the Ideal Gas Law equation to solve for the volume:
V1 = nRT1 / P
V2 = nRT2 / P
Substituting the given values, we have:
V1 = n * 0.0821 * 288.15 / 1
V2 = n * 0.0821 * 303.15 / 1
2. The Change in Internal Energy:
The change in internal energy (ΔU) can be calculated using the equation:
ΔU = (3/2) * n * R * (T2 - T1)
Substituting the given values, we have:
ΔU = (3/2) * n * 0.0821 * (303.15 - 288.15)
3. The Work Done by Helium:
The work done by helium can be calculated as the area under the curve on a P-V diagram. In this case, since the pressure is constant, the work done can be simplified to:
Work = P * (V2 - V1) = 1 * (V2 - V1)
Substituting the values, we have:
Work = V2 - V1
4. The Change in Enthalpy:
The change in enthalpy (ΔH) can be calculated from the change in internal energy and the work done:
ΔH = ΔU + P * ΔV
Since the pressure is constant, the term P * ΔV simplifies to Work. Therefore:
ΔH = ΔU + Work
Substituting the values, we have:
ΔH = ΔU + V2 - V1
With the equations provided, you can now calculate the values for each of the quantities.
To answer these questions, we can utilize the ideal gas law and the equations related to thermodynamics. Let's go through each question one by one:
1. The initial and final volume:
According to the ideal gas law, PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature in Kelvin. Initially, the temperature is 15 degrees Celsius, which is equal to 15 + 273 = 288 Kelvin. The pressure is given as 1.00 atm.
1 mole of any gas occupies 22.4 liters at standard temperature and pressure (STP). Therefore, for each mole of helium, the initial volume can be calculated as follows:
V1 = (n1 * R * T1) / P = (1 * 0.0821 * 288) / 1.00 = 23.6 liters
Similarly, using the final temperature of 30 degrees Celsius (30 + 273 = 303 Kelvin), we can calculate the final volume as:
V2 = (n2 * R * T2) / P = (1 * 0.0821 * 303) / 1.00 = 24.8 liters
So, the initial volume is 23.6 liters, and the final volume is 24.8 liters.
2. The change in internal energy:
The internal energy (ΔU) of an ideal gas is only dependent on temperature and is given by the equation ΔU = (n * Cv * ΔT), where ΔT is the change in temperature and Cv is the molar heat capacity at constant volume. For helium, Cv is approximately 3R/2.
Using the given temperature change (ΔT = T2 - T1 = 30 - 15 = 15 degrees Celsius = 15 Kelvin), we can calculate the change in internal energy as:
ΔU = (n * Cv * ΔT) = (1 * 3 * 0.0821/2 * 15) = 1.236 Joules/mol
Therefore, the change in internal energy is approximately 1.236 Joules/mol.
3. The work done by helium:
Since the pressure remains constant throughout the process, the work done by the helium gas can be calculated using the formula W = PΔV.
ΔV = V2 - V1 = 24.8 - 23.6 = 1.2 liters
Converting the volume change to liters to cubic meters (1 liter = 0.001 m^3), we get:
ΔV = 1.2 * 0.001 = 0.0012 m^3
Hence, the work done by helium is:
W = PΔV = (1.00 atm) * (0.0012 m^3) = 1.20 J
Thus, the work done by helium is approximately 1.20 Joules.
4. The change in enthalpy:
The change in enthalpy (ΔH) can be calculated using the equation ΔH = ΔU + PΔV. We have already calculated ΔU and ΔV in the previous answers.
ΔH = (n * Cv * ΔT) + PΔV = 1.236 + (1.00 atm) * (0.0012 m^3) = 1.236 + 1.20 = 2.44 Joules/mol
Therefore, the change in enthalpy is approximately 2.44 Joules/mol.