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Question #43948

Difficulty: 8/10 | 🟢 Free

The Question

An ideal gas of molar mass $M$ flows steadily and isothermally at a constant temperature $T$ through a horizontal converging-diverging nozzle. At the inlet, the cross-sectional area is $A_1$ and the gas velocity is $v_1$ . At the outlet, the cross-sectional area is $A_2$ and the gas velocity is $v_2$ . Assuming ideal gas behavior and neglecting viscous effects, which of the following equations correctly relates the velocities and areas? (Here, $R$ is the universal gas constant).

Options

A.
$\ln\left( \frac{A_1 v_1}{A_2 v_2} \right) = \frac{M(v_1^2 - v_2^2)}{2RT}$
B.
$A_1 v_1 = A_2 v_2$
C.
$\ln\left( \frac{A_1 v_1}{A_2 v_2} \right) = \frac{M(v_2^2 - v_1^2)}{2RT}$
D.
$\frac{A_1 v_1}{A_2 v_2} = 1 + \frac{M(v_1^2 - v_2^2)}{2RT}$

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Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Multi-Concept_Bridge

Master problems that require linking physics and math concepts to solve complex competitive questions.

Concepts Behind This Question

Isothermal Processes|Bernoulli’s Principle|Small Angle Approximation

Why This Question Is Tricky

Common Mistakes

Students often blindly apply the standard Bernoulli's equation ( $P + \frac{1}{2}\rho v^2 = \text{constant}$ ) and the incompressible continuity equation ( $A_1 v_1 = A_2 v_2$ ), forgetting that these are only valid for incompressible fluids (constant density).

In this isothermal gas flow, density is directly proportional to pressure ( $\rho = PM/RT$ ).

The trap tests whether students recognize the need to couple the differential form of Euler's equation ( $dP = -\rho v dv$ ) with the ideal gas law and the compressible continuity equation ( $\rho_1 A_1 v_1 = \rho_2 A_2 v_2$ ) to derive the correct non-linear velocity-area relationship.

Option D represents the trap of applying incompressible Bernoulli to find the pressure difference and then using compressible continuity.

Quick Facts About This Question

Question Information

Subject	Physics
Chapter	Kinetic Theory - Class 11 Part 2
Concept	Adiabatic/Isothermal Processes via KTG
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free