Cooling Curve of a Sealed Gas Sample
A rigid 2.0 L steel vessel contains 0.020 mol of nitrogen (diatomic, ideal) at 300 K. A student heats it to 600 K using a hot plate, then removes it and lets it cool in air at 293 K. During heating, the pressure gauge reads 0.25 MPa at 300 K and 0.50 MPa at 600 K. The student claims the gas did no work, so all heat went into internal energy, and computes Q = n Cv ΔT with Cv = (5/2)R. A classmate argues that because the gas is diatomic, the vibrational modes must also be excited at 600 K and the heat capacity should be (7/2)R. The vessel's pressure rating is 1.0 MPa.
1. Assess the student's claim that no work is done and that Q = n Cv ΔT is the correct heat input.
Read the worked answer
For a rigid sealed vessel, volume is constant, so W = ∫P dV = 0. The first law then gives Q = ΔU, and for an ideal gas U depends only on T, so ΔU = n Cv ΔT. The student's conclusion is correct, but the reasoning must be stated carefully: Q = n Cv ΔT is valid here because V is constant, not as a general rule. If the vessel were flexible, Q = n Cp ΔT would apply instead.
Numerically: ΔU = 0.020 * (5/2)(8.314) * 300 = 0.020 * 20.785 * 300 ≈ 125 J. So Q ≈ 125 J into the gas during heating, and about the same out during cooling.
2. Evaluate the classmate's argument about vibrational modes and decide which heat capacity is appropriate at 600 K.
Read the worked answer
The classmate is wrong for this temperature range. N2 has a vibrational characteristic temperature of about 3390 K, so at 600 K the vibrational modes are essentially frozen (kT << the vibrational spacing). Equipartition applies only to modes that are thermally accessible.
At 300-600 K, N2 has 3 translational and 2 rotational degrees of freedom active, giving Cv = (5/2)R. The (7/2)R value would require fully excited vibration, which needs temperatures on the order of thousands of kelvin. So the student's Cv = (5/2)R is the right choice, and the classmate's correction would overestimate Q by 40%.
3. Using the pressure readings, check whether the gas behaves ideally and whether the vessel is safe at 600 K.
Read the worked answer
From the ideal gas law at 300 K: P = nRT/V = 0.020 * 8.314 * 300 / 0.0020 = 49.9/0.0020 ≈ 2.49e4 Pa = 0.025 MPa. But the gauge reads 0.25 MPa, a factor of 10 higher. That is a red flag: either the amount of gas is 0.20 mol, the volume is 0.20 L, or the gauge is misread. The ratio P(600)/P(300) = 0.50/0.25 = 2.0, which matches T(600)/T(300) = 2.0 and is consistent with ideal behavior in the ratio, so the gas is ideal-like but the absolute numbers are inconsistent.
Safety: even at the stated 0.50 MPa reading, the vessel is below its 1.0 MPa rating. But if the true amount is 10x larger, the pressure at 600 K would be about 5.0 MPa, far above the rating. The student must resolve the discrepancy before continuing; the safe action is to stop and recheck the gas quantity and gauge units.