Starting at temperature 300 K, one mole of an
ideal diatomic gas ($\gamma$ = 1.4) is first compressed
adiabatically from volume V1 to V2 = ${{{V_1}} \over {16}}$. It is
then allowed to expand isobarically to volume
2V2. If all the processes are the quasi-static then
the final temperature of the gas (in oK) is (to
the nearest integer) _____.
Answer (integer)
1818
Solution
T<sub>1</sub>V<sub>1</sub><sup>$\gamma$–1</sup> = T<sub>2</sub>V<sub>2</sub>
<sup>$\gamma$–1</sup>
<br><br>$$300 \times {V^{{7 \over 5} - 1}} = {T_2}{\left( {{V \over {16}}} \right)^{{7 \over 5} - 1}}$$
<br><br>$\Rightarrow$ T<sub>2</sub> = 300 × (16)<sup>0.4</sup>
<br><br>Isobaric process
<br><br>V = ${{nRT} \over P}$
<br><br>V<sub>2</sub>
= kT<sub>2</sub>... (1)
<br><br>2V
<sub>2</sub>
= KT<sub>f</sub>... (2)
<br><br>T<sub>f</sub> = 2T<sub>2</sub> = 300 × 2 × (16)<sup>0.4</sup>
= 1818 K
About this question
Subject: Physics · Chapter: Thermodynamics · Topic: Zeroth and First Law
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