Solved: A rigid, perfectly insulated container has a

Chapter 18, Problem 10DQ

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QUESTION:

Problem 10DQ

A rigid, perfectly insulated container has a membrane dividing its volume in half. One side contains a gas at an absolute temperature T0 and pressure p0, while the other half is completely empty. Suddenly a small hole develops in the membrane, allowing the gas to leak out into the other half until it eventually occupies twice its original volume. In terms of T0 and p0, what will be the new temperature and pressure of the gas when it is distributed equally in both halves of the container? Explain your reasoning.

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QUESTION:

Problem 10DQ

ANSWER:

Solution 10DQ

Introduction

Since the system is perfectly insulated, the expansion will be a adiabatic process. We will first use the adiabatic relationship between P and V and then calculate the final pressure. From the final pressure and volume we will then calculate the final temperature.

Step 1

Suppose the initial volume of the gas, that is the volume of the half of the container is ${V}_{0}$ . So the final volume of the gas is $2{V}_{0}$ .

From thermodynamics, we know that, for adiabatic process, the relationship between P and V is given by

$P{V}^{\gamma{}}=constant$

$\Rightarrow{}{P}_{1}{{V}_{1}}^{\gamma{}}={P}_{2}{{V}_{2}}^{\gamma{}}$

Here $\gamma{}$ is the specific heat ratio of the gas and is given by $\gamma{}=\frac{{C}_{P}}{{C}_{v}}$ . Where ${C}_{p}$ and ${C}_{v}$ are the specific heat of the gas at constant volume and constant pressure.

Here we have ${V}_{1}={V}_{0}$ and ${V}_{2}=2{V}_{0}$ and ${P}_{1}={P}_{0}$ . Using this values we have

${P}_{2}=\frac{{P}_{1}{{V}_{1}}^{\gamma{}}}{{{V}_{2}}^{\gamma{}}}=\frac{{P}_{0}{{V}_{0}}^{\gamma{}}}{(2{{V}_{0}}^{\gamma{}})}=\frac{{P}_{0}}{{2}^{\gamma{}}}$