1yaar see.....
T1/T2=(V2/V1)γ-1....
if u take γ=1.66 i.e. for a monoatomic gas....
T1/T2=(V2/V1)0.66
and since V1>V2.....
V2/V1<1.....
THIS IMPLIES.....
(V2/V1)0.66<1
in adiabatic compression decrease in volume is assosiated with
a)inc in temp n decre in pressure
b)decr in temp n incre in pressr
c)decre in both
d0inc in both
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19 Answers
1
1gamma always lies between 1 and 2 only never equal to both due to its relation with degree of freedom.
its D
AS IN ADIABATIC COMPRESSION PRESSURE INCREASES AND TEMPERATURE ALSO INCREASES
MORE LIKE A HEATER
BUT NOT EXACTLY
1in that case we cann't be damn sure about the answer for sure....
1bt how cud u be sure fr all d gases i mean monoatomic hasmax value ofγ if n ki value be 1000000000000000then
1arrey yaar suppose γ=1.123
γ-1=.123
frraction ki power 0.123 can be greater thn 1 na i hpe u understand[1]
1bt how u saw γ-1
i mean v2/v1is a fraction
n if suppose γ=1. sumting then d ans ll be reversed
1yaar pressure bhi toh increase hoga.....
V is decreasing....
PVγ=constant.....
1P1V1γ = P2V2γ
P1V1γ / P2V2γ = 1
P1V1 / P2V2 = (V2/V1)γ-1
T1/T2 = (V2/V1)γ-1
V2/V1 is a fraction ...... so T2 final temp is more than initial and haan volume is decreasing so pressure increases ... oops i did a silly mistake !.
ans is D