A container at an initial volume of 0.07 m^3 holds 1 mole ot a monatomic ideal...

Question

Question

Answer 1

Answer #1

In adiabatic process

PV^(^alpha) = constant

alpha = 1.66 for an ideal monoatomic gas

so,

P1V1^(^alpha) = P2V2^(^alpha)

1*(0.07/0/03)^1.66 = P2 = 4.08 atm

V2T2^(1/1.66-1) = V1 T1(1/1.66-1)

T2 = 611.9 k

Answer 2

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