JEE Advanced · 2010 · 11 Apr 2010 · Paper 2 · Heat & Thermodynamics

A diatomic ideal gas is compressed adiabatically to

\frac{1}{32}

of its initial volume. In the initial temperature of the gas is

T_{i}

(in Kelvin) and the final temperature is

a T_{i}

, the value of

a

Answer: 4

Solution

For an adiabatic process:

T V^{γ - 1} = constant

For a diatomic gas,

γ = \frac{7}{5}

.

Therefore:

T_{i} V_{i}^{2/5} = T_{f} V_{f}^{2/5}

T_{i} V_{i}^{2/5} = a T_{i} (\frac{V _{i}}{32})^{2/5}

1 = a \times (32)^{- 2/5} = a \times (2^{5})^{- 2/5} = a \times 2^{- 2}

a = 4

The answer is 4.

JEE Advanced · 2010 · 11 Apr 2010 · Paper 2 · Integer answer

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