Question

The volume of a gas is reduced adiabatically to 1/4 of its volume at 27°C. If γ = 1.4 the new temperature is(a) $(300) 2^{0.4}\ K$ (b) $(300) 2^{1.4}\ K$(c) $300 (4)^{0.4}\ K$(d) $300 (2)^{0.4}\ K$

Answer 1

Hey brainly user

Here is your answer

Given,

V1=V

V2=V/4

T1=300K

T2=?

Formula :

$\frac{t1}{t2} = ( \frac{v2}{v1} ) {}^{ \gamma - 1}$

$\frac{300}{t2} = ( \frac{ \frac{v}{4} }{v} ) {}^{ \gamma - 1}$

$\frac{300}{t2} = ( \frac{1}{4} ) {}^{0.4}$

$t2 = 300 \times 4 {}^{0.4}$