The volume of an ideal gas increases 8 times and temperature becomes (1/4 )^ th of initial temperature during a reversible change. If there is no exchange of heat in this process ( Q=0 ) then identif…

Question

The volume of an ideal gas increases 8 times and temperature becomes (1/4 )^ th of initial temperature during a reversible change. If there is no exchange of heat in this process ( Q=0 ) then identify the gas from the following options (Assuming the gases given in the options are ideal gases):

Accepted Answer

Correct answer: D. (d) For an adiabatic process ( Q=0 ) with an ideal gas: TV^ -1= constant Given: V_f=8V_i and T_f= 14T_i Applying the adiabatic relation: T_iV_i^ -1=T_fV_f^ -1 T_iV_i^ -1= T_i4 (8V_i )^ -1 1= 14 8^ -1 4=8^ -1 2^2=2^3( -1) 2=3( -1) = 53=1.67 This value corresponds to a monatomic gas. Among the options, only He is monatomic ( =5/3 for monatomic ideal gases). O_2 and N_2 are diatomic ( =7/5 ), CO_2 is polyatomic ( ≈1.30), and NH_3 is polyatomic.

$The volume of an ideal gas increases 8 times and temperature becomes (1/4)^{th} of initial temperature during a reversible change. If there is no exchange of heat in this process (Δ Q = 0) then identify the gas from the following options (Assuming the gases given in the options are ideal gases):$

Solution

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