Question

One mole of an ideal gas at an initial temperature of
T K does 6R joules of work adiabatically. If the ratio
of specific heats of this gas at constant pressure and a constant volume is 5/3, the final temperature of gas will be​

Answer 1

$\bf \to \: \green{{ \underline{given \div }}}$

$\sf \to \: one \: moles \: of \: an \: ideal \: gas \\ \\ \sf \to \: initial \: temperature \: \: = t \: k \\ \\ \sf \to \: work \: done \: = 6r \: joule \: adiabatically \\ \\ \sf \to \: specific \: heat \: of \: the \: gas \: at \: constant \: pressure \: and \: volume \: = \dfrac{5}{3}$

$\bf \to \: \pink{ { \underline{to \: find \: final \: temperature \: of \: gas \: }}}$

$\bf \to \: \orange{{ \underline{step \: - by - \: step \: - explanation}}}$

$\sf \to \: { \underline{in \: adiabatic \: process \: }} \\ \\ \sf \to \: the \: system \: is \: insulated \: in \: surrounding \\ \sf \to \: and \: heat \: absorb \: or \: released \: is \: = 0 \\ \\ \sf \to \: pv {}^{( \lambda)} = constant \\ \\ \sf \to \: { \underline{work \: done \: by \: gas \: in \: adiabatic \: process}} \\ \\ \sf \to \: w \: = \dfrac{ \mu \: r \: ( t_{i} \: - \: t_{f} \: ) }{( \gamma \: - 1)} \\ \\ \sf \to \: 6r \: = \dfrac{1 \times r \: (t \: - t_{f} \: ) }{ (\dfrac{5}{3} - 1)} \\ \\ \sf \to \: 6 = \frac{(t \: - \: t_{f}) }{ \frac{2}{3} } \\ \\ \sf \to \: 4 = t \: - \: t_{f} \\ \\ \sf \to \: \green{{ \underline{t \: =( t_{f} \: - 4 )k}}}$

Answer 2

$\sf \large \red{\underline{ Question:-}}$

• One mole of an ideal gas at an initial temperature of T K does 6R joules of work adiabatically. If the ratio of specific heats of this gas at constant pressure and a constant volume is 5/3, the final temperature of gas will be

$\\\\\sf \large \red{\underline{Given:-}}$

• $\sf \: t_{1} = t$

• $\sf w \: = 6r$

• $\sf \: \gamma = \frac{5}{3}$

$\\\\\sf \large \red{\underline{To \: Find:-}}$

• $\sf \: the \: final \: temperature \: of \: gas \:will \: be$

$\\\\\sf \large \red{\underline{Solution :- }}$

$\sf \underline {we \: have \: adiabatic \: formula : }$

$\boxed{ \sf \red {\: w = \frac{r( t_{2} - t_{1})}{ 1 - \gamma } }}$

$\sf \underline { \orange {putting \: all \: values}}$

$\sf \to \: 6r \: = \frac{r( t_{2} - t)}{ 1 - \frac{5}{3} } \\ \\ \sf \to \: 6r = \frac{r(t_{2} - t)}{ \frac{3 - 5}{3} } \\ \\ \sf \to 6r = \frac{r( t_{2} -t) }{ \frac{ - 2}{3} } \\ \\ \sf \to 6 \cancel{r} = \frac{ \cancel{r}( t_{2} - t)}{ \frac{ - 2}{3} } \\ \\ \sf \to \: t_{2} - t = 6 \times \frac{ (- 2)}{3} \\ \\ \sf \to \: t_{2} - t = \frac{ - 12}{3} \\ \\ \sf \to \: t_{2} - t \: = \frac{ - \cancel{ 12}}{ \cancel{3}} \\ \\ \sf \to \: t_{2} - t = - 4 \\ \\ \sf \to \red{ t_{2} = (t - 4)k }$

$\\ \sf \text{ \underline{\sf \red{the final temperature of gas will be}}} (t - 4)k \: \huge \dag$