11.Thermodynamics
medium

A diatomic gas initially at $18^o C$ is compressed adiabatically to one-eighth of its original volume. The temperature after compression will be

A

${10^o}C$

B

${887^o}C$

C

$668K$

D

${144^o}C$

(AIPMT-1996)

Solution

$T{V^{\gamma – 1}} = $ constant

$ \Rightarrow {T_2} = {T_1}{\left( {\frac{{{V_1}}}{{{V_2}}}} \right)^{\gamma – 1}} $$= (273 + 18)\;{\left( {\frac{V}{{V/8}}} \right)^{0.4}} = 668\;K$

Standard 11
Physics

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