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JIPMER 2015
Exam
JIPMER
Year
2015
Subject
Physics
Chapter
Kinetic Theory of Gases

JIPMER 2015 · Physics · Kinetic Theory of Gases

A horizontal tube of length l closed at both ends, contains an...

A horizontal tube of length ll closed at both ends, contains an ideal gas of molecular weight MM. The tube is rotated at a constant angular velocity ω\omega about a vertical axis passing through an end. Assuming the temperature to be uniform and constant. If p1p_1 and p2p_2 denote the pressure at free and the fixed end respectively, then choose the correct relation.A horizontal tube of length l closed at both ends, contains an ideal gas of molecular weight M . The tube is rotated at

Options

  1. Ap2p1=eMω2l22RT\frac{p_2}{p_1}=e^{\frac{M \omega^2 l^2}{2 R T}}
  2. Bp1p2=eMων2RT\frac{p_1}{p_2}=e^{\frac{M \omega \nu^2}{R T}}
  3. Cp1p2=eω/M3RT\frac{p_1}{p_2}=e^{\frac{\omega / M}{3 R T}}
  4. Dp2p1=eM2ω2I23RT\frac{p_2}{p_1}=e^{\frac{M^2 \omega^2 I^2}{3 R T}}

Correct answer

A. p2p1=eMω2l22RT\frac{p_2}{p_1}=e^{\frac{M \omega^2 l^2}{2 R T}}

Step-by-step solution

Consider the diagram A horizontal tube of length l closed at both ends, contains an ideal gas of molecular weight M . The tube is rotated at Consider the elemetary part of thickness dxd x at a distance xx from axis of rotation, then force on this part Adp=(dm)ω2x…A d p=(d m) \omega^2 x \ldots (i) where, dm=d m= mass of element Now, from ideal gas equation pV=nRTp V=n R T we get pAdx=dmMRT⇒dm=MpARTdxp A d x=\frac{d m}{M} R T \Rightarrow \quad d m=\frac{M p A}{R T} d x ...(ii) From Eqs. (i) and (ii) Adp=MpARTω2xdx∫p1p2dpp=∫01Mω2RTxdx⇒ln⁡p2p1=Mω2I22RT⇒p2p1=eMω2I22RT\begin{aligned} & A d p=\frac{M p A}{R T} \omega^2 x d x \\ & \int_{p_1}^{p_2} \frac{d p}{p}=\int_0^1 \frac{M \omega^2}{R T} x d x \\ & \Rightarrow \quad \ln \frac{p_2}{p_1}=\frac{M \omega^2 I^2}{2 R T} \\ & \Rightarrow \quad \frac{p_2}{p_1}=e^{\frac{M \omega^2 I^2}{2 R T}} \\ & \end{aligned}
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