You have 1L of an ideal gas at 0 degree celsius and 10 atm pressure. You allow the gas to expand against a constant external pressure of 1 atm, while the temperature remains constant. Assuming, 101.3 J/liter-atm, find q,w,delta E, and delta H in Joules, A) if the expansion took place in a vacuum and b) if the gas were expanded to 1 atm pressure.
To solve this problem, we can use the equations for work (w), heat (q), internal energy (ΔE), and enthalpy (ΔH). Using the ideal gas equation PV = nRT, where P is pressure, V is volume, n is the number of moles, R is the ideal gas constant, and T is temperature, we can calculate the volume change (ΔV) during the expansion.
Given:
Initial pressure (P1) = 10 atm
Final pressure:
A) In vacuum: P2 = 0 atm
B) At 1 atm: P2 = 1 atm
Initial volume (V1) = 1 L
Temperature (T) = 0 °C = 273.15 K
External Pressure (Pext):
A) In vacuum: Pext = 0 atm
B) At 1 atm: Pext = 1 atm
Conversion factor: 1 atm x 101.3 J/L-atm = 101.3 J
Step 1: Calculating the volume change (ΔV)
Using the ideal gas equation PV = nRT, we can rearrange it to solve for volume:
V = nRT / P
Given:
P1 = 10 atm
V1 = 1 L
T = 273.15 K
R (ideal gas constant) = 0.08206 L·atm/(mol·K) (or you can use 8.314 J/(mol·K))
Plugging in the values:
V1 = (1 mol x 0.08206 L·atm/(mol·K) x 273.15 K) / 10 atm
Calculating V1:
V1 = 2.246 L
For A) in vacuum:
Volume change (ΔV) = V2 - V1 = 0 - 2.246 L = -2.246 L
For B) at 1 atm:
Using the ideal gas equation with P2 = 1 atm:
V2 = (1 mol x 0.08206 L·atm/(mol·K) x 273.15 K) / 1 atm
Calculating V2:
V2 = 2.794 L
Volume change (ΔV) = V2 - V1 = 2.794 L - 2.246 L = 0.548 L
Step 2: Calculating work (w)
The work done in an expansion against a constant external pressure is given by:
w = -Pext ΔV
Given:
Pext:
A) In vacuum: Pext = 0 atm (= 0 atm x 101.3 J/L-atm = 0 J/L)
B) At 1 atm: Pext = 1 atm (= 1 atm x 101.3 J/L-atm = 101.3 J/L)
ΔV:
A) In vacuum: ΔV = -2.246 L
B) At 1 atm: ΔV = 0.548 L
For A) in vacuum:
w = 0 J/L x (-2.246 L) = 0 J
For B) at 1 atm:
w = 101.3 J/L x (0.548 L) = 55.47 J
Step 3: Calculating heat (q)
For an expansion or compression of an ideal gas, the heat is given by:
q = ΔE + w
Since the expansion takes place at constant temperature, the change in internal energy (ΔE) is zero.
For A) in vacuum:
q = ΔE + w = 0 J + 0 J = 0 J
For B) at 1 atm:
q = ΔE + w = 0 J + 55.47 J = 55.47 J
Step 4: Calculating change in internal energy (ΔE) and enthalpy (ΔH)
Since ΔE = q - w and temperature remains constant, ΔE and ΔH will have the same values as q.
For A) in vacuum:
ΔE = 0 J
ΔH = 0 J
For B) at 1 atm:
ΔE = 55.47 J
ΔH = 55.47 J
Therefore, the results are:
A) If the expansion took place in a vacuum:
- q = 0 J
- w = 0 J
- ΔE = 0 J
- ΔH = 0 J
B) If the gas were expanded to 1 atm pressure:
- q = 55.47 J
- w = 55.47 J
- ΔE = 55.47 J
- ΔH = 55.47 J
To find q, w, ΔE, and ΔH, we can use the first law of thermodynamics, which states that the change in internal energy (ΔE) of a system is equal to the heat transferred (q) minus the work done (w) on the system.
a) If the expansion took place in a vacuum:
Since the expansion is taking place in a vacuum, there is no external pressure to overcome, so the work done (w) is zero.
q = ΔE
w = 0 (no work is done)
ΔH = q (assuming constant pressure)
To find q (heat transferred), we can use the equation:
q = n * R * ΔT
where n is the number of moles of gas, R is the ideal gas constant (0.0821 L·atm/(mol·K)), and ΔT is the change in temperature.
Given that we have 1 L of gas, we can calculate the number of moles using the ideal gas law:
PV = nRT
n = PV / RT
where P is the pressure, V is the volume, T is the temperature, and R is the ideal gas constant.
Let's plug in the values:
P = 10 atm
V = 1 L
T = 0 °C = 273.15 K
R = 0.0821 L·atm/(mol·K)
n = (10 atm * 1 L) / (0.0821 L·atm/(mol·K) * 273.15 K)
n ≈ 0.443 mol
Now, we can calculate q:
q = n * R * ΔT
Since the temperature remains constant, ΔT = 0.
q = 0.443 mol * 0.0821 L·atm/(mol·K) * 0 K
q = 0 J
Therefore, q = 0 J, w = 0 J, ΔE = 0 J, and ΔH = 0 J.
b) If the gas were expanded to 1 atm pressure:
In this case, the external pressure is 1 atm, so the work done (w) can be calculated using the equation:
w = -Pext * ΔV
where Pext is the external pressure and ΔV is the change in volume.
Given that the volume increases from 1 L to an unknown final volume, we can substitute ΔV with Vfinal - Vinitial.
Let's calculate w:
w = -Pext * ΔV
Pext = 1 atm
Vinitial = 1 L
Vfinal = unknown
To find the final volume, we need to use the Boyle's law, which states that for a given amount of gas at constant temperature, the product of pressure (P) and volume (V) is constant.
Pinitial * Vinitial = Pfinal * Vfinal
Pinitial = 10 atm
Pfinal = 1 atm
Vinitial = 1 L
Vfinal = unknown
Using the equation, we can solve for Vfinal:
(10 atm * 1 L) = (1 atm * Vfinal)
Vfinal = 10 L
Now we have all the values to calculate w:
w = -Pext * ΔV
w = -(1 atm) * (Vfinal - Vinitial)
w = -(1 atm) * (10 L - 1 L)
w = -9 atm * L
w = -9 L·atm
The negative sign indicates work done by the system.
Now let's calculate q and ΔE using the first law of thermodynamics:
q = ΔE + w
ΔE = q - w
We already found ΔE to be 0 J in part a).
Now, q = ΔE + w
q = 0 J + (-9 L·atm) = -9 L·atm
ΔH = q (assuming constant pressure)
ΔH = -9 L·atm
Finally, we can convert ΔH from L·atm to Joules using the given conversion factor:
1 L·atm = 101.3 J
ΔH = -9 L·atm * 101.3 J/L·atm
ΔH = -911.7 J
Therefore, q = -9 L·atm, w = -9 L·atm, ΔE = 0 J, and ΔH = -911.7 J.