Answer:
(a) 0.203 moles
(b) 900 K
(c) 900 K
(d) 15 L
(e) A → B, W = 0, Q = Eint = 1,518.91596 J
B → C, W = Q ≈ 1668.69974 J Eint = 0 J
C → A, Q = -2,531.5266 J, W = -1,013.25 J, Eint = -1,518.91596 J
(g) ∑Q = 656.089 J, ∑W = 655.449 J, ∑Eint = 0 J
Explanation:
At point A
The volume of the gas, V₁ = 5.00 L
The pressure of the gas, P₁ = 1 atm
The temperature of the gas, T₁ = 300 K
At point B
The volume of the gas, V₂ = V₁ = 5.00 L
The pressure of the gas, P₂ = 3.00 atm
The temperature of the gas, T₂ = Not given
At point C
The volume of the gas, V₃ = Not given
The pressure of the gas, P₃ = 1 atm
The temperature of the gas, T₂ = T₃ = 300 K
(a) The ideal gas equation is given as follows;
P·V = n·R·T
Where;
P = The pressure of the gas
V = The volume of the gas
n = The number of moles present
R = The universal gas constant = 0.08205 L·atm·mol⁻¹·K⁻¹
n = PV/(R·T)
∴ The number of moles, n = 1 × 5/(0.08205 × 300) ≈ 0.203 moles
The number of moles in the sample, n ≈ 0.203 moles
(b) The process from points A to B is a constant volume process, therefore, we have, by Gay-Lussac's law;
P₁/T₁ = P₂/T₂
∴ T₂ = P₂·T₁/P₁
From which we get;
T₂ = 3.0 atm. × 300 K/(1.00 atm.) = 900 K
The temperature at point B, T₂ = 900 K
(c) The process from points B to C is a constant temperature process, therefore, T₃ = T₂ = 900 K
(d) For a constant temperature process, according to Boyle's law, we have;
P₂·V₂ = P₃·V₃
V₃ = P₂·V₂/P₃
∴ V₃ = 3.00 atm. × 5.00 L/(1.00 atm.) = 15 L
The volume at point C, V₃ = 15 L
(e) The process A → B, which is a constant volume process, can be carried out in a vessel with a fixed volume
The process B → C, which is a constant temperature process, can be carried out in an insulated adjustable vessel
The process C → A, which is a constant pressure process, can be carried out in an adjustable vessel with a fixed amount of force applied to the piston
(f) For A → B, W = 0,
Q = Eint = n·cv·(T₂ - T₁)
Cv for monoatomic gas = 3/2·R
∴ Q = 0.203 moles × 3/2×0.08205 L·atm·mol⁻¹·K⁻¹×(900 K - 300 K) = 1,518.91596 J
Q = Eint = 1,518.91596 J
For B → C, we have a constant temperature process
Q = n·R·T₂·㏑(V₃/V₂)
∴ Q = 0.203 moles × 0.08205 L·atm/(mol·K) × 900 K × ln(15 L/5.00 L) ≈ 1668.69974 J
Eint = 0
Q = W ≈ 1668.69974 J
For C → A, we have a constant pressure process
Q = n·Cp·(T₁ - T₃)
∴ Q = 0.203 moles × (5/2) × 0.08205 L·atm/(mol·K) × (300 K - 900 K) = -2,531.5266 J
Q = -2,531.5266 J
W = P·(V₂ - V₁)
∴ W = 1.00 atm × (5.00 L - 15.00 L) = -1,013.25 J
W = -1,013.25 J
Eint = n·Cv·(T₁ - T₃)
Eint = 0.203 moles × (3/2) × 0.08205 L·atm/(mol·K) × (300 K - 900 K) = -1,518.91596 J
Eint = -1,518.91596 J
(g) ∑Q = 1,518.91596 J + 1668.69974 J - 2,531.5266 J = 656.089 J
∑W = 0 + 1668.69974 J -1,013.25 J = 655.449 J
∑Eint = 1,518.91596 J + 0 -1,518.91596 J = 0 J
. From the figure we know that:
