Two moles of an ideal monoatomic gas occupies | vedclass

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Question

In an adiabatic process

$T{V^{\gamma  - 1}}=constant$   or    ${T_1}{V_1}^{\gamma  - 1} = {T_2}{V_2}^{\gamma  -

Vedclass · Accepted Answer

$(a)$ $189$ $K$       $(b)$ $-2.7$ $kJ$

Vedclass · Answer

In an adiabatic process

$T{V^{\gamma  - 1}}=constant$   or    ${T_1}{V_1}^{\gamma  - 1} = {T_2}{V_2}^{\gamma  - 1}$

For monoatomic gas    $\gamma  = \frac{5}{3}$

$\left( {300} \right){V^{2/3}} = {T_2}{\left( {2V} \right)^{2/3}} \Rightarrow {T_2} = \frac{{300}}{{{{\left( 2 \right)}^{2/3}}}}$

${T_2} = 189\,K\,\left( {final\,temperature} \right)$

Change in internal energy $\Delta U = n\frac{f}{2}R\,\Delta T$

$ = 2\left( {\frac{3}{2}} \right)\left( {\frac{{25}}{3}} \right)\left( { - 111} \right) =  - 2.7\,kJ$

Two moles of an ideal monoatomic gas occupies a volume $V$ at $27^o C$. The gas expands adiabatically to a volume $2\ V$. Calculate $(a)$ the final temperature of the gas and $(b)$ change in its internal energy.

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