A balloonist is preparing to make a trip in a helium-filled balloon. The trip begins in early morning at a temperature of 15 degrees C. By midafternoon, the temperature has increased to 30. degrees C. Assuming the pressure remains constant at 1.00atm, for each mole of helium, calculate:
(a) The initial and final volumes
(b) The change in internal energy, ΔE [Hint: Helium behaves like an ideal gas, so E=3/2nRT. Be sure the units of R are consistent with those of E.]
(c) The work (w) done by the helium (in J)
(d) The heat (q) transferred (in J)
(e) ΔH for the process (in J)
(f) Explain the relationship between the answers to (d) and (e).
c) work = -p*delta v
d) q = mass x specific heat x (Tfinal-Tinitial)
e) qp = delta H
a)Pv=nRT
1mol(.0821)(288.15)/1atm=23.7L
1mol(.0821)(303.15).1atm=24.9L
b)E=3/2nRT
1mol(8.314R)(303.15-288.15)=187J
oshili onda nyenwa lela
(a) Well, if the pressure remains constant and the temperature increases, we can use Charles's Law to find the initial and final volumes. Charles's Law states that the volume of a gas is directly proportional to its temperature at constant pressure. So, we have:
V1/T1 = V2/T2
Plugging in the values:
V1/15 = V2/30
Solving for V1, we get:
V1 = (15/30) * V2
V1 = 0.5V2
So, the initial volume is half of the final volume.
(b) The change in internal energy, ΔE, can be calculated using the formula E = (3/2)nRT. Since we are given that the pressure remains constant at 1.00 atm, we can rewrite the equation as:
ΔE = (3/2)nRΔT
Where ΔT is the change in temperature. Since we are going from 15 degrees C to 30 degrees C, ΔT = 30 - 15 = 15 degrees C.
(c) The work done by the helium can be calculated using the formula w = -PΔV, where P is the pressure and ΔV is the change in volume. Since the pressure remains constant at 1.00 atm, we can rewrite the equation as:
w = -P(V2 - V1)
Substituting V1 = 0.5V2, we get:
w = -P(0.5V2 - V2)
w = -0.5PV2
(d) The heat transferred, q, can be calculated using the equation q = ΔE + w. Substituting the values we calculated earlier:
q = ΔE + w
q = (3/2)nRΔT - 0.5PV2
(e) ΔH for the process is equal to the heat transferred, q, since the pressure is constant.
ΔH = q
ΔH = (3/2)nRΔT - 0.5PV2
(f) The relationship between the answers to (d) and (e) is that they are equal. ΔH represents the change in enthalpy, which is equal to the heat transferred at constant pressure. In this case, the pressure remains constant, so ΔH = q.
To solve this problem, we'll use the ideal gas law and the equation for the internal energy of an ideal gas. Let's go step by step:
(a) The initial and final volumes:
We know that the pressure remains constant (1.00 atm). 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.
To calculate the initial volume, we convert the initial temperature from Celsius to Kelvin:
Initial Temperature (T1) = 15°C + 273.15 = 288.15 K
Using PV = nRT, we can rearrange the equation to solve for V:
V1 = (nRT1) / P = (n * 0.0821 L.atm/mol.K * 288.15 K) / 1.00 atm
Similarly, we calculate the final volume using the final temperature in Kelvin:
Final Temperature (T2) = 30°C + 273.15 = 303.15 K
V2 = (nRT2) / P = (n * 0.0821 L.atm/mol.K * 303.15 K) / 1.00 atm
(b) The change in internal energy, ΔE:
The equation for the internal energy of an ideal gas is given by E = (3/2)nRT, where E is the internal energy, n is the number of moles, R is the ideal gas constant, and T is the temperature in Kelvin.
ΔE = E2 - E1 = ((3/2)nRT2) - ((3/2)nRT1) = (3/2)nR(T2 - T1)
(c) The work done by the helium (w):
The work done by the helium can be calculated using the equation w = -PΔV, where w is the work done, P is the pressure, and ΔV is the change in volume.
ΔV = V2 - V1
w = -P(ΔV) = -P(V2 - V1)
(d) The heat transferred (q):
The heat transferred can be calculated using the first law of thermodynamics, which states that ΔE = q + w.
q = ΔE - w
(e) ΔH for the process:
ΔH is the change in enthalpy and can be calculated using the equation ΔH = ΔE + PΔV. Since we are assuming constant pressure in this problem, ΔH simplifies to ΔH = ΔE.
(f) The relationship between (d) and (e):
For this process where pressure is constant, ΔH is equal to the change in internal energy (ΔE). So the answers to (d) and (e) would be the same.