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.

Question

Quizplus · Accepted Answer

From the diagram, we can see that the process from state 1 to state 2 is an isothermal compression, so we can use the ideal gas law with constant temperature to find the initial pressure:
P1V1 = nRT1
P1 = nRT1/V1

Since V2 = V1/4, we know that P2 = 4P1.

Now we can use the ideal gas law with constant volume to find the final pressure and temperature for state 3:
P2/3 = T2/T3
P3 = P2T3/T2
nRT3/V2 = 4nRT2/V1
T3 = 4T2V2/V1

Similarly, we can use the ideal gas law with constant pressure to find the final volume and temperature for state 4:
V3/4 = T3/T4
V4 = V3T4/T3
nRT4/P3 = nRT1/P1
T4 = P3T1/T3

Plugging in the given values, we get:
 P1 = (4.3 g / 32 g/mol) * (0.0821 L atm/mol K) * (21 + 273 K) / 10 L = 2 atm
 P2 = 4P1 = 8 atm
 T2 = 21 + 273 = 294 K
 T3 = 4T2V2/V1 = 4(294 K)(10 L) / 2.5 L = 1180 K
 T4 = P3T1/T3 = (2 atm)(21 + 273 K) / (8 atm) = 57 K

Converting the temperatures to Celsius, we get:
 T3 = 1180 - 273 = 907°C
 T4 = 57 - 273 = -216°C

Therefore, the answer is A) -52°C, 390°C (rounded to the nearest 10°C).

The Figure Shows a PV Diagram for 4