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

Using the ideal gas law, we can find the initial pressure of the gas in state 1:
P1V1 = nRT1
P1 = (nRT1)/V1 = [(4.3 g/16 g/mol) x (0.0821 L∙atm/mol∙K) x (21°C + 273.15)]/(2.5 L) = 2.51 atm

From the diagram, we can see that the pressure decreases to 1 atm in state 2 (isothermal expansion) and then increases to 3 atm in state 3 (isobaric compression) before decreasing back to 2.51 atm in state 4 (isothermal compression).
Therefore, we have:
P2 = P4 = 1 atm
P3 = 3 atm

To find the temperatures of states 3 and 4, we can use the ideal gas law again:
For state 3:
P3V3 = nRT3
T3 = (P3V3)/(nR) = (3 atm x 2.5 L)/[(4.3 g/16 g/mol) x (0.0821 L∙atm/mol∙K)] = 660°C

For state 4:
P4V4 = nRT4
T4 = (P4V4)/(nR) = (1 atm x 2.5 L)/[(4.3 g/16 g/mol) x (0.0821 L∙atm/mol∙K)] = -52°C (Note: this is below 0°C, so it is expressed as a negative value)

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

The Figure Shows a PV Diagram for 4