0.1 gm molecule of diatomic gas is isothermally expanded at 293 K… - Vidyadip

Question

0.1 gm molecule of diatomic gas is isothermally expanded at 293 K to twice its initial volume. After that its adiabatic expansion is done until the final volume becomes four times the initial volume. Find the final temperature of the gas and also calculate the work done for both the expansion ( $R = 8 . 3$ joule $m o l ^{ l } K ^{-1}$, $\gamma=1.4)$

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Answer

$n=0.1 gm$ molecule of diatomic gas
In an isothermal :
$V_1=V \text { and } T_1=293 K$
In the begining $\quad V _2=2 V$ and $T _2=293 K$
In adiabatic process
$\begin{array}{rlrl} V _{ i } & =2 V, & T _i & =293 K \\ V _{ f } & =4 V, & T _{ f } & =? \\ \gamma & =1.4 & \end{array}$
For change in adiabatic process:
$\begin{aligned} T _{ i } V _{ i }^{\gamma-1} & = T _{ f } V _{ f }^{\gamma-1} \\ \left(\frac{V_{ i }}{ V _{ f }}\right)^{\gamma-1} & =\frac{ T _{ f }}{ T _{ i }}\end{aligned}$
Putting the value :
$\begin{aligned}\left(\frac{2 V}{4 V}\right)^{1,4-1} & =\frac{T_r}{293} \\\left(\frac{1}{2}\right)^{0.4} & =\frac{T_3}{293}\end{aligned}$
Taking $\log$ both sides
$\log \left(\frac{1}{2}\right)^{0.4}=\log \frac{T_2}{293}$
$\begin{array}{lrl}\Rightarrow & 0.4(\log 1-\log 2) & =\log T _2-\log 293 \\ \Rightarrow & 0.4(0-0.3010) & =\log T _2-2.4669 \\ \Rightarrow & -0.1204+2.4669 & =\log T _2 \\ \Rightarrow & 2.3465 & =\log T _2\end{array}$
$\begin{array}{l}\therefore\quad\quad T _2=\text { Antilog }\{2.3465\} \\\quad\quad\quad T _2=222.1 K \end{array}$
Formula of work done in isothernal process :
$W =2.303 \times nRT \log _{10} \frac{V_2}{V_1}$
Putting the value
$ \begin{aligned} W & =2.303 \times 0.1 \times 8.3 \times 293 \log _{10}\left(\frac{2 V}{V}\right) \\ W & =560.07 \times \log 2 \\ & =560.07 \times 0.3010=168.58 \\ W & =168.58 \text { Joule } \end{aligned} $
Now its adiabatic expansion is done until the final volumne becomes four times the initial volume. Work done is adiabatic process :
$\begin{aligned}\text {Putting the value}\quad W & =\frac{0.1 \times 8.3}{1-1.4}[222.1-293] \\ & =\frac{0.1 \times 8.3 \times 70.9}{0.4}=147.12 \text {Joule} \\ W & =147.12 \text { Joule }\end{aligned}$

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