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 (ΔQ = 0) then identify the gas from the following options (Assuming the gases given in the options are ideal gases) :
Options

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 (ΔQ = 0) then identify the gas from the following options (Assuming the gases given in the options are ideal gases) :

Options

G1 Physics · Accepted Answer

Because there is no heat exchange and the process is reversible, the process is adiabatic. For an ideal gas in a reversible adiabatic process, T × [V raised to the power (γ-1)] = constant. Let the initial state be T₁, V₁. The final state is T₂ = T₁/4, V₂ = 8V₁. Therefore, T₁ × [V₁ raised to the power (γ-1)] = (T₁/4) × [(8V₁) raised to the power (γ-1)]. Cancel the common factor involving T₁ and V₁: 1 = (1/4) × [8 raised to the power (γ-1)]. Thus, 4 = 8 raised to the power (γ-1). Since 4 = 2² and…