Question

A cylinder contains 0.100 mol of an ideal monatomic gas. Initially the gas is at$\mathbf{1}\mathbf{.}\mathbf{00}\mathbf{\times }{\mathbf{10}}^{\mathbf{5}}\mathbf{Pa}$and occupies a volume of$\mathbf{2}\mathbf{.}\mathbf{50}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{3}}{\mathbf{m}}^{\mathbf{-}\mathbf{3}}$. (a) Find the initial temperature of the gas in kelvins. (b) If the gas is allowed to expand to twice the initial volume, find the final temperature (in kelvins) and pressure of the gas if the expansion is (i) isothermal; (ii) isobaric; (iii) adiabatic.

Short answer

(a) By using the ideal gas equation The initial temperature of gas is ${\mathrm{T}}_1=301\mathrm{K}$

(b) (i) For Isobaric process the final temperature and pressure is ${\mathrm{T}}_2=602\mathrm{K}$, ${\mathrm{p}}_2=1.00\mathrm{x}10^5\mathrm{Pa}$Respectively.

(ii) For Isobaric process the final temperature and pressure is${\mathrm{T}}_2=602\mathrm{K}$, ${\mathrm{p}}_2=1.00\mathrm{x}{10}^5\mathrm{Pa}$Respectively.

(iii) For adiabatic process the final temperature and pressure is ${\mathrm{T}}_2=189\mathrm{K}$, ${\mathrm{p}}_2=3.14\mathrm{x}{10}^4\mathrm{Pa}$Respectively.

Step-by-step solution

Calculate the initial temperature of gas

By using the ideal gas equation,

pV = nRT…..(1)

Where = n Number of moles, R = Universal gas constant$\left[\mathrm{R}=8.314\frac{\mathrm{J}}{\mathrm{mol}.\mathrm{k}}\right]$ , V = Volume of gas,

p = Pressure, and T = Temperature of gas

For calculate the initial temperature

${\mathrm{T}}_1=\frac{{\mathrm{p}}_1{\mathrm{V}}_1}{\mathrm{nR}}$

We were given an ideal monoatomic gas with a number of moles n = 0.100mol and Occupies a volume${\mathrm{V}}_1=2.5\times {10}^{-3}{\mathrm{m}}^3$ at the initial pressure${\mathrm{p}}_1=1.00\times {10}^5\mathrm{Pa}$ .

Substitute all given value to calculate${\mathrm{T}}_1$

role="math" localid="1668176715829" $\begin{array}{r}{\mathrm{T}}_1=\frac{1.00\times {10}^5\mathrm{Pa}\times 2.5\times {10}^{-3}{\mathrm{m}}^3}{0.100\mathrm{mol}\times 8.314\frac{\mathrm{J}}{\mathrm{mol}.\mathrm{k}}}\\ {\mathrm{T}}_1=301\mathrm{K}\end{array}$

Hence, The initial temperature of gas is${\mathrm{T}}_1=301\mathrm{K}$

Step 2:

1. Here, We have to find the Final temperature${\mathrm{T}}_2$ and final pressure . If the initial volume is twice the final volume${\mathrm{V}}_2,\left({\mathrm{V}}_2=2{\mathrm{AV}}_1\right)$ , At given three processes. Isothermal, isobaric, and Adiabatic.

Step 2(i) : For Isothermal process

For any system. If the temperature is constant. During the process Then this process is known as isothermal process. So. In this process, the initial temperature is equal to final temperature ${\mathrm{T}}_1={\mathrm{T}}_2$.

So,

${\mathrm{T}}_2=301\mathrm{K}$

From equation first

pV = nRT

We can establish the relation between initial state and final state because Is constant.

${\mathrm{p}}_1{\mathrm{V}}_1={\mathrm{p}}_2{\mathrm{V}}_2$

We know that , $\left({\mathrm{V}}_2=2{\mathrm{V}}_1\right)$and ${\mathrm{p}}_1=1.00\times {10}^5\mathrm{Pa}$substitute these values and calculate

$\begin{array}{r}{\mathrm{p}}_2\\ {\mathrm{p}}_2={\mathrm{p}}_1\left(\frac{{\mathrm{V}}_1}{{\mathrm{V}}_2}\right)\\ {\mathrm{p}}_2=1.00\times {10}^5\mathrm{Pa}\left(\frac{{\mathrm{V}}_1}{2{\mathrm{V}}_1}\right)\\ {\mathrm{p}}_2=5\times {10}^4\mathrm{Pa}\end{array}$

Therefor, for isothermal process the final temperature and pressure is${\mathrm{T}}_2=301\mathrm{K},{\mathrm{p}}_2=5\times {10}^4\mathrm{Pa}$ ,Respectively.

Step 2(ii) : For isobaric process

For isobaric process. The pressure is constant In the process.

${\mathrm{p}}_1={\mathrm{p}}_2$

So,

${\mathrm{p}}_2=1.00\times {10}^5\mathrm{Pa}$

From equation first , To calculate the value of ${\mathrm{T}}_2$.

$\begin{array}{r}\mathrm{pV}=\mathrm{nRT}\\ \frac{\mathrm{V}}{\mathrm{T}}=\frac{\mathrm{nR}}{\mathrm{p}}\end{array}$

Also term $\frac{\mathrm{nR}}{\mathrm{p}}$is constant So, We can establish the relation between initial state and final state

$\begin{array}{r}\frac{{\mathrm{V}}_1}{{\mathrm{T}}_1}=\frac{{\mathrm{V}}_2}{{\mathrm{T}}_2}\\ {\mathrm{T}}_2=\frac{{\mathrm{T}}_1{\mathrm{V}}_2}{{\mathrm{V}}_1}\\ \mathrm{We}\mathrm{know}\mathrm{that}\left({\mathrm{V}}_2=2{\mathrm{V}}_1\right),\mathrm{and}{\mathrm{T}}_1=301\mathrm{K}\mathrm{substitute}\mathrm{these}\mathrm{values}\mathrm{we}\mathrm{get}\\ {\mathrm{T}}_2=301\mathrm{K}\left(\frac{2{\mathrm{V}}_1}{{\mathrm{V}}_1}\right)\\ {\mathrm{T}}_2=602\mathrm{K}\end{array}$

Therefor, for Isobaric process the final temperature and pressure is ${\mathrm{T}}_2=602\mathrm{K}$ ,${\mathrm{p}}_2=1.00\times {10}^5\mathrm{Pa}$ Respectively.

Step 2(ii) : For adiabatic process

An adiabatic process in which no heat transfer takes place between a system and its surroundings. Zero heat transfer is an idealization.

${\mathrm{TV}}^{\mathrm{Y}-1}=\mathrm{Constant}$

Where, y = The ratio of heat capacities$\left[\mathrm{Y}=\frac{{\mathrm{C}}_{\mathrm{p}}}{{\mathrm{C}}_{\mathrm{V}}}\right]$

Thus for an initial state$\left({\mathrm{T}}_1{\mathrm{V}}_1\right)$and a final state$\left({\mathrm{T}}_2{\mathrm{V}}_2\right)$

${\mathrm{T}}_{!}{{\mathrm{V}}_1}^{\mathrm{Y}-1}$……(2) (adiabatic process, ideal gas)

For monoatomic gas${\mathrm{C}}_{\mathrm{V}}=\frac{3\mathrm{R}}{2}$

And

$\begin{array}{r}{C}_p=C_v+R\\ C_p=\frac{3R}{2}+R\\ C_p=\frac{5R}{2}\end{array}$

So,

$\begin{array}{r}\mathrm{Y}=\frac{{\mathrm{C}}_{\mathrm{P}}}{{\mathrm{C}}_{\mathrm{V}}}\\ \mathrm{Y}=\frac{5\mathrm{R}/2}{3\mathrm{R}/3}\\ \mathrm{Y}=1.67\\ \mathrm{Now}\mathrm{put}\mathrm{all}\mathrm{the}\mathrm{values}\mathrm{in}\mathrm{equation}2\mathrm{nd}\\ {\mathrm{T}}_2={\mathrm{T}}_1{\left(\frac{{\mathrm{V}}_1}{{\mathrm{V}}_2}\right)}^{\mathrm{r}-1}\\ ={\mathrm{T}}_1{\left(\frac{{\mathrm{V}}_1}{2{\mathrm{V}}_1}\right)}^{1.67-1}\\ =301\mathrm{K}{\left(\frac{1}{2}\right)}^{187-1}\\ {\mathrm{T}}_2=189\mathrm{K}\end{array}$

Now we have to calculate So, from ideal gas equation Also

localid="1668312614887" $$\begin{gathered} V_2=2V_1=2\times 2.5\times {10}^{-3}{\mathrm{m}}^3 \\ {\mathrm{p}}_2{\mathrm{V}}_2={\mathrm{nRT}}_2 \\ {\mathrm{p}}_2=\frac{{\mathrm{nRTT}}_2}{{\mathrm{V}}_2} \\ =\frac{0.100\mathrm{mol}\times 8.314\mathrm{J}/{\mathrm{mol}}^{-\mathrm{K}}\times 189\mathrm{K}}{5\times {10}^{-3}{\mathrm{m}}^3} \\ {\mathrm{p}}_2=3.14\times {10}^4\mathrm{Pa} \end{gathered}$$

Hence, for adiabatic process the final temperature and pressure is ${\mathrm{T}}_2=189\mathrm{K}$,${\mathrm{p}}_2=3.14\times {10}^4\mathrm{Pa}$Respectively.