An ideal gas is trapped inside a test tube of | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) In the first case,the pressure of the gas $P_1$ is given by $P_1 = P_0 + \frac{Mg}{A}$.In the second case,when the tube is inverted,the pressure o

Vedclass · Accepted Answer

$\frac{101}{99}$

Vedclass · Answer

(B) In the first case,the pressure of the gas $P_1$ is given by $P_1 = P_0 + \frac{Mg}{A}$.In the second case,when the tube is inverted,the pressure of the gas $P_2$ is given by $P_2 = P_0 - \frac{Mg}{A}$.Since the process is isothermal,$P_1 V_1 = P_2 V_2$. Since $V = A 	imes L$,we have $P_1 L_1 = P_2 L_2$.Therefore,$\frac{L_2}{L_1} = \frac{P_1}{P_2} = \frac{P_0 + \frac{Mg}{A}}{P_0 - \frac{Mg}{A}} = \frac{P_0 A + Mg}{P_0 A - Mg}$.Substituting the values: $P_0 A = 10^5 	imes 20 	imes 10^{-6} = 2 \, N$ and $Mg = 0.002 	imes 10 = 0.02 \, N$.$\frac{L_2}{L_1} = \frac{2 + 0.02}{2 - 0.02} = \frac{2.02}{1.98} = \frac{202}{198} = \frac{101}{99}$.

Similar Questions

The condition $dQ = dW$ holds good in which of the following?

In an isothermal change,the change in pressure and volume of a gas can be represented for three different temperatures $T_3 > T_2 > T_1$ as:

Write the first law of thermodynamics for an isothermal process in an ideal gas.

$Assertion :$ The isothermal curves intersect each other at a certain point.
$Reason :$ The isothermal change takes place slowly,so,the isothermal curves have very little slope.

In isothermal expansion,the pressure is determined by

Vedclass Test Series

Exam Paper Generator

Online Exam Module