An ideal gas undergoes a process in which its density ρ is proportional to T 2 1 , and the molar heat capacity in this process is 4 R . Then:

V ∝ T 2 and P ∝ V T ∝ T 1 , so P V 2 1 = constant and x = 2 1 . W = 1 − x R Δ T = 2 R Δ T . C = C v + 2 R = 4 R gives C v = 2 R (4 degrees of freedom), C p = 3 R , γ = 2 3 . A, B, D.

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PhysicsHeat & ThermoThermodynamics Laws & Specific Heats of GasesHard2 minQB

PhysicsHardmultiple choice

An ideal gas undergoes a process in which its density $ρ$ is proportional to $\frac{1}{T ^{2}}$ , and the molar heat capacity in this process is $4 R$ . Then:

Options:(select one or more)

Answer:

A, B, D

Solution:

$V \propto T^{2}$ and $P \propto \frac{T}{V} \propto \frac{1}{T}$ , so $P V^{\frac{1}{2}} = constant$ and $x = \frac{1}{2}$ . $W = \frac{R Δ T}{1 - x} = 2 R Δ T$ . $C = C_{v} + 2 R = 4 R$ gives $C_{v} = 2 R$ (4 degrees of freedom), $C_{p} = 3 R$ , $γ = \frac{3}{2}$ . A, B, D.

Subject:PhysicsTopic:Heat & ThermoSubtopic:Thermodynamics Laws & Specific Heats of Gases

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