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

From the diagram, we can see that the process from state 1 to state 2 is an isobaric process (constant pressure), since the volume increases while the pressure stays constant. Therefore, we can use the following equation to find the temperature of state 2:

P1V1/T1 = P2V2/T2

where P1V1/T1 is the initial condition (state 1), P2V2 is the final condition (state 2), and T2 is the temperature we are solving for. Plugging in the values, we get:

(1 atm)(0.0100 L)/(294 K) = (1.5 atm)(0.0100 L)/T2

Simplifying, we get:

T2 = (1.5 atm)(294 K)/(1 atm) = 441 K = 168°C

Next, we can see that the process from state 2 to state 3 is an isochoric process (constant volume), since the volume stays the same while the pressure increases. Therefore, we can use the following equation to find the temperature of state 3:

P2/T2 = P3/T3

where P2/T2 is the initial condition (state 2), P3/T3 is the final condition (state 3), and T3 is the temperature we are solving for. Plugging in the values, we get:

(1.5 atm)/(441 K) = (3 atm)/T3

Simplifying, we get:

T3 = (3 atm)(441 K)/(1.5 atm) = 882 K = 609°C

Finally, we can see that the process from state 3 to state 4 is an isobaric process (constant pressure), since the volume decreases while the pressure stays constant. Therefore, we can use the following equation to find the temperature of state 4:

P3V3/T3 = P4V4/T4

where P3V3/T3 is the initial condition (state 3), P4V4 is the final condition (state 4), and T4 is the temperature we are solving for. Plugging in the values, we get:

(3 atm)(0.00250 L)/(882 K) = (1.5 atm)(0.00250 L)/T4

Simplifying, we get:

T4 = (1.5 atm)(882 K)/(3 atm) = 441 K = 168°C

Therefore, the answer is A) -52°C, 390°C.

The Figure Shows a PV Diagram for 4