The figure shows a pV diagram for 4.3 g of oxygen gas (O₂) in a sealed container. The temperature T₁ of the gas in state 1 is 21°C. What are the temperatures T₃ and T₄ of the gas in states 3 and 4? The ideal gas constant is R = 8.314 J/mol ∙ K = 0.0821 L ∙ atm/mol ∙ K, and the ATOMIC weight of oxygen is 16 g/mol. A) -52°C, 390°C B) 16°C, 47°C C) 220°C, 660°C D) 11°C, 32°C

Question

Quizplus · Accepted Answer

Using the ideal gas law, we can find the pressure and volume for each state.

State 1: 
V1 = nRT1/P1 = (4.3 g)/(16 g/mol) x (0.0821 L∙atm/mol∙K) x (294 K)/(1 atm) = 0.915 L 
P1 = 2.5 atm

State 2: 
V2 = V1 = 0.915 L 
P2 = 1.5 atm

State 3: 
P3 = P2 = 1.5 atm 
V3 = nRT3/P3 = (4.3 g)/(16 g/mol) x (0.0821 L∙atm/mol∙K) x (T3)/(1.5 atm)

State 4: 
V4 = V3 = (0.915 L)(1.5 atm)/(2.5 atm) = 0.549 L 
P4 = 2.5 atm

We can now solve for T3:

T3 = P3V3/nR = (1.5 atm)(0.915 L)/(0.2693 mol)(0.0821 L∙atm/mol∙K) = 220°C

To find the temperature of state 4, we can use the fact that the process from state 3 to state 4 is an isochoric (constant volume) process, so:

(P3/T3) = (P4/T4)

T4 = P4T3/P3 = (2.5 atm)(220 K)/(1.5 atm) = 367 K = 94°C

Converting to Celsius:

State 3: 220°C 
State 4: 94°C = -52°C (rounded to nearest degree)

Therefore, the answer is choice A.

The Figure Shows a PV Diagram for 4