Chapter 6, Problem 6.77QP

(a)

Interpretation Introduction

Interpretation:

The final temperature of the given sample has to be calculated.

Concept Introduction:

The ideal gas Law equation is,

$\begin{array}{l}\text{PV}\text{=}\text{nRT}\\ \text{P}\to \text{Pressure}\\ \text{V}\to \text{Volume}\\ \text{n}\to \text{Number}\text{of}\text{moles}\\ \text{R}\to \text{Ideal}\text{gas}\text{constant}\\ \text{T}\to \text{Temperature}\end{array}$

The first law of thermodynamics states that energy can be either destroyed or created but instead it can be converted from one form to other.

The equation for the first law of thermodynamics can be given as,

  $\text{ΔU=q+w}$

$W=-PdV$ ; Where P is the pressure, dV is the change in volume.

$\begin{array}{c}\text{ΔU}\text{=}{\text{C}}_{\text{v}}\text{dT}\\ \text{q}\text{=}{\text{C}}_{\text{P}}\text{dT}\\ R\text{=}{\text{C}}_{\text{p}}{\text{-C}}_{\text{v}}\end{array}$

Here $C_{p}and{C}_v$ are the molar heat capacity at constant pressure and molar heat capacity at constant volume respectively.

Explanation of Solution

Given,

A $\text{1}\text{.00}\text{-}\text{mole}$ sample of ammonia at $14.0atm$ and ${25}^{0}C$ in a cylinder fitted with a movable piston expands against a constant external pressure of $1.00atm$. At equilibrium, the pressure and volume of the gas are $1.00atmand23.5L$ , respectively.

The final temperature of the sample can be calculated as follows,

$\begin{array}{c}\text{P}\text{=}\text{14}\text{atm}\\ \text{V=}\text{?}\\ T{\text{=25}}^{\text{o}}\text{C}\end{array}$

The volume of the substance will change in accordance to the piston’s movement.

Thus,

According to the Ideal gas equation,

$\begin{array}{c}V=\frac{nRT}{P}\\ =\frac{1\times 0.0821\times 298}{14}\\ =1.75L\end{array}$

At equilibrium the volume is $23.5L$ thus, $dV=23.5-1.75=21.75L$ ,

$\begin{array}{c}\text{W}\text{=}\text{-Pdv}\\ \text{=}\text{-}\left(\text{1×21}\text{.75}\right)\text{L}\\ \text{=}\text{-21}\text{.75}\text{atm}\text{.L}\end{array}$

 $atm.L=101.33J$

Therefore,

$\begin{array}{c}\text{W}\text{=}\text{-21}\text{.75}\text{atm}\text{.L}\text{×}\text{101}\text{.33}\text{J}\\ \text{=}\text{-2204}\text{J}\end{array}$

First law of thermodynamics can be written as,

$\begin{array}{c}\text{Δ}\text{U}\text{=}\text{q}\text{+}\text{W}\\ {\text{C}}_{\text{v}}\text{dT}\text{=}{\text{C}}_{\text{p}}\text{dT}\text{+}\text{w}\\ \left({\text{C}}_{\text{p}}{\text{-C}}_{\text{v}}\right)\text{dT}\text{=}\text{-W}\\ \text{dT}\text{=}\frac{\text{-W}}{\text{R}}\end{array}$

Substituting the known values in the above equation,

$\begin{array}{c}\text{dT}=\frac{\text{2204}\text{J}}{\text{8}\text{.314}\text{J/K}}\\ =\text{265K}\end{array}$

$\begin{array}{c}\text{dT}\text{=}{\text{T}}_{\text{Final}}\text{-}{\text{T}}_{\text{Initial}}\\ \text{265K}\text{=}{\text{T}}_{\text{Final}}\text{-}\text{298}\text{K}\\ {\text{T}}_{\text{Final}}\text{=}\text{265}\text{K+298}\text{K}\\ {\text{T}}_{\text{Final}}\text{=}\text{563}\text{K}\end{array}$

Therefore, final temperature (${\text{T}}_{\text{Final}}$) of the given sample is $\text{563}\text{K}$

(b)

Interpretation Introduction

Interpretation:

$q,wand\Delta U$ for the given process have to be determined.

Concept Introduction:

Specific heat:

Specific heat can be defined as quantity of heat required to raise the temperature of $\text{1}\text{g}$ substance by $\text{1°C}$.  The relationship between heat and change in temperature can be expressed by the equation given below.

    $\text{q}=\text{cmΔT}$

Where   $\text{q=}$ Heat added

       c= Specific heat

    $\text{m=}$ Mass

  $\text{ΔT=}$ Change in temperature.

The first law of thermodynamics states that energy can be either destroyed or created but instead it can be converted from one form to other.

The equation for the first law of thermodynamics can be given as,

  $\text{ΔU=q+w}$

$W=-PdV$ ; Where P is the pressure, dV is the change in volume.

$\begin{array}{c}\text{ΔU}\text{=}{\text{C}}_{\text{v}}\text{dT}\\ \text{q}\text{=}{\text{C}}_{\text{P}}\text{dT}\\ R\text{=}{\text{C}}_{\text{p}}{\text{-C}}_{\text{v}}\end{array}$

Here $C_{p}and{C}_v$ are the molar heat capacity at constant pressure and molar heat capacity at constant volume respectively.

Explanation of Solution

A $\text{1}\text{.00}\text{-}\text{mole}$ sample of ammonia at $14.0atm$ and ${25}^{0}C$ in a cylinder fitted with a movable piston expands against a constant external pressure of $1.00atm$. At equilibrium, the pressure and volume of the gas are $1.00atmand23.5L$ , respectively.

At equilibrium the volume is $23.5L$ thus,$dV=23.5-1.75=21.75L$,

$\begin{array}{c}\text{W}\text{=}\text{-Pdv}\\ \text{=}\text{-}\left(\text{1×21}\text{.75}\right)\text{L}\\ \text{=}\text{-21}\text{.75}\text{atm}\text{.L}\end{array}$

 $atm.L=101.33J$

Therefore,

$\begin{array}{c}\text{W}\text{=}\text{-21}\text{.75}\text{atm}\text{.L}\text{×}\text{101}\text{.33}\text{J}\\ \text{=}\text{-2204}\text{J}\end{array}$

First law of thermodynamics can be written as,

$\begin{array}{c}\text{Δ}\text{U}\text{=}\text{q}\text{+}\text{W}\\ {\text{C}}_{\text{v}}\text{dT}\text{=}{\text{C}}_{\text{p}}\text{dT}\text{+}\text{w}\\ \left({\text{C}}_{\text{p}}{\text{-C}}_{\text{v}}\right)\text{dT}\text{=}\text{-W}\\ \text{dT}\text{=}\frac{\text{-W}}{\text{R}}\end{array}$

Substituting the known values in the above equation,

$\begin{array}{c}\text{dT}=\frac{\text{2204}\text{J}}{\text{8}\text{.314}\text{J/K}}\\ =\text{265K}\end{array}$

$\begin{array}{c}\text{q}=\text{cmΔT}\\ \text{molar}\text{mass}\text{of}\text{ammonia}=17.03\text{mol}\text{/g}\\ \text{c}=\text{0}\text{.0258}\text{J/g}{\text{.}}^{\text{0}}\text{C}\\ \text{ΔT}=\text{265}\text{K}\text{=}\text{-8}\text{.05}{}^{\text{0}}\text{C}\\ \text{q}=17.\text{03}\text{mol}\text{/g}\text{×}\text{0}\text{.0258}\text{J/g}{\text{.}}^{\text{0}}\text{C}\text{×}\left(\text{-8}\text{.15}{}^{\text{o}}\text{C}\right)\\ =\text{-3}\text{.54}{\text{J}}^{\text{o}}\text{C}\\ \end{array}$

The equation for the first law of thermodynamics is,

  $\text{ΔU=q+w}$

$\Delta U$ can be determined as follows,

$\begin{array}{l}q=-3.54J^{o}C\\ W=-2204J\end{array}$

$\begin{array}{c}\text{ΔU=}\left(\text{-3}\text{.54}\right)\text{+}\left(\text{-2204}\right)\\ \\ \text{=}-2207.5J\end{array}$

For the given process,

$W=-2204J$

$\text{q}\text{=}\text{-3}\text{.54}{\text{J}}^{\text{o}}\text{C}$

$\text{ΔU}\text{=-2207}\text{.5}\text{J}$