The pressure and density of a diatomic gas (γ=75) changes adiabatically from (p,d) to(p1,d1). If d′d=32 then p′p is
A
1128
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B
32
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C
128
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D
256
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Solution
The correct option is B 128 in an adiabatic process relation between P and V is : PVγ=constant and V=Md so Pdγ=constant P1dγ1=P2dγ2 so P′P=(d′d)γ P′P=(132)75 P′P=128