A diatomic ideal gas is compressed adaibatically to 1/32 of its initial volume. If the initial temperature of the gas is `T_i` (in Kelvin) and the final temperature is a `T_i`, the value of a is

A diatomic ideal gas is compressed adaibatically to 1/32 of its initia

1.	A diatomic ideal gas is compressed adaibatically to 1/32 of its initial volume. If the initial temperature of the gas is `T_i` (in Kelvin) and the final temperature is a `T_i`, the value of a is
Answer» Correct Answer - D For an adiabatic process, the temperature-volume relationship is `T_1V_1^(gamma-1)=T_2V_2^(gamma-1) rArr T_1=T_2((V_2)/(V_1))^(gamma-1)` Here `gamma=1.4 (for diatomic gas). V_2=(V_1)/32, T_1=T_i, T_2=aT_i` `:. T_i=aT_i[1/32]^(1.4-1) =aT_i[1/(2^5)]^0.4=(aT_i)/4 :. a=4`

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A diatomic ideal gas is compressed adaibatically to 1/32 of its initial volume. If the initial temperature of the gas is `T_i` (in Kelvin) and the final temperature is a `T_i`, the value of a is

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