Question

Diatomic gas at pressure P and volume V is compressed adiabatically to 1/32 times the original volume. Then the final pressure is?​

Answer 1

Answer:

for adiabatic process:-

$p {v}^{ \gamma } = constant$

and,

gamma for diatomic gas=7/5

so,

$p1 {v1}^{ \gamma } = p2 {v2}^{ \gamma }$

v2 is 1/32 times of v1.

[p1=initial pressure]

[p2=final pressure]

[v2=initial volume]

[v2=final volume]

$p2 = p1 { \frac{v1}{v2} }^{ \gamma }$. ............(1)

and,

$\frac{v1}{v2} = 32$

${32}^{7 \div 5} = {2}^{7} = 128$. ................(2)

so,from equation (1) and equation (2)

p2=128p1.

According to the question, the pressure will become 128 times the initial pressure P.

plZ MaRK as BrAinLIesT.

HOPE THAT HELPS.

Answer 2

Answer:

128 times than the original volume