A certain mass of gas at 273 K is expanded to | vedclass

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Question

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

Vedclass · Accepted Answer

$-182$

Vedclass · Answer

(b) For adiabatic process $T{V^{\gamma - 1}}$= 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{1}{{81}}} ight)^{1.25 - 1}} 	imes 273$$ = {\left( {\frac{1}{{81}}} ight)^{0.25}} 	imes 273$

$ = \frac{{273}}{3} = 91K = \,-182&deg;C$

A certain mass of gas at $273 K$ is expanded to $81$ times its volume under adiabatic condition. If $\gamma = 1.25$ for the gas, then its final temperature is ..... $^oC$

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