The volume of a gas is reduced adiabatically | vedclass

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Question

(b) For adiabatic change $T{V^{\gamma - 1}}$= constant
==> $\frac{{{T_2}}}{{{T_1}}} = {\left( {\frac{{{V_1}}}{{{V_2}}}} \right)^{\gamma - 1}}$

Vedclass · Accepted Answer

$300 	imes {4^{0.4}}K$

Vedclass · Answer

(b) For adiabatic change $T{V^{\gamma - 1}}$= constant
==> $\frac{{{T_2}}}{{{T_1}}} = {\left( {\frac{{{V_1}}}{{{V_2}}}} ight)^{\gamma - 1}}$

==> ${T_2} = {\left( {\frac{{{V_1}}}{{{V_2}}}} ight)^{\gamma - 1}} 	imes {T_1}$
==> ${T_2} = {\left( {\frac{V}{{V/4}}} ight)^{1.4 - 1}} 	imes 300$ $ = 300 	imes {(4)^{0.4}}K$

The volume of a gas is reduced adiabatically to $\frac{1}{4}$ of its volume at $27°C$, if the value of $\gamma = 1.4,$ then the new temperature will be

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