A rigid diatomic ideal gas is compressed adiabatically to (1/32) of its initial volume. If the initial temperature of the gas is T_1 (in kelvin) and the final temperature is " nT_1 ", the value of n is \ γ= 1.4\

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Correct answer: D. 4

For an adiabatic process, the relation between temperature and volume is given by:

T_1 V_1^{\gamma - 1} = T_2 V_2^{\gamma - 1}

Given that

V_2 = \dfrac{V_1}{32}

,

T_2 = nT_1

, and

\gamma = 1.4

, we can substitute these values into the equation:

T_1 V_1^{1.4 - 1} = nT_1 \left(\dfrac{V_1}{32}\right)^{1.4 - 1}

V_1^{0.4} = n \left(\dfrac{V_1}{32}\right)^{0.4}

1 = n \left(\dfrac{1}{32}\right)^{0.4}

n = (32)^{0.4}

Since

0.4 = \dfrac{2}{5}

and

32 = 2^5

, we have:

n = (2^5)^{\dfrac{2}{5}} = 2^2 = 4

Answer:

4

A rigid diatomic ideal gas is compressed adiabatically to (1/32) of...

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