Chapter 3, Problem 3.27P

(a)

Interpretation Introduction

Interpretation:

Calculate the final temperature of a rigid non conducting tank with volume of $\text{4}{\text{m}}^{\text{3}}$ after the removal of thin membrane separates it intotwo unequal subparts by ratio $2:3$ using a thin membrane, representing $1/3$ of tank contains nitrogen gas at temperature ${\text{100}}^{\text{o}}\text{C, 6 bar}$ and other $2/3$ of tank is evacuated. How much work is done and is this process reversible?

Concept Introduction:

According to first law of thermodynamics

  $dU=\delta Q+\delta W$

Where $dU$ is change in internal energy, $\delta Q$ is heat transfer and $\delta W$ is external workdone.

And $dU=C_{V}dt$

(b)

Interpretation Introduction

Interpretation:

On the same process if an amount of work is applied on membrane by which the gas can be returned to its initial state to make a reversible process. Calculate the work is done.

Concept Introduction:

Since temperature remains constant $dT=0$ this means isothermal process. For isothermal process work is done on reversible process is given by

  $W=-P_2{V}_2\ln (\frac{V_2}{V_1})=-P_2{V}_2\ln (\frac{P_1}{P_2})$............(1)