Starting at temperature 300 K , one mole of an ideal , diatomic gas γ=1.4 is first compressed adiabatically from volume V1 to V2=V116 . 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 K ) is (to the nearest integer)___

During adiabatic processTVγ−1=constant ⇒T1V1γ−1=T2.V2γ−1 ⇒T2=T1(V1V2)γ−1=300(16)75−1=T2=300(24)2/5 =T2=300×24×25During isobaric processV=nRTP ⇒V∝T ⇒V2V3=T2T3 ⇒V22V2=300×24×25T3 ⇒T3=2×300×28/5 =1818.85 =1819 K

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Starting at temperature $300 K$ , one mole of an ideal , diatomic gas $(γ = 1.4)$ is first compressed adiabatically from volume $V_{1}$ to $V_{2} = \frac{V_{1}}{16}$ . It is then allowed to expand isobarically to volume $2 V_{2}$ . If all the processes are the quasi-static then the final temperature of the gas (in $K$ ) is (to the nearest integer)___

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answer is 1819.

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Detailed Solution

During adiabatic process

$\begin{array}{l} T V^{γ - 1} = c o n s \tan t \\ \Rightarrow T_{1} {V_{1}}^{γ - 1} = T_{2} . V_{2}^{γ - 1} \\ \Rightarrow T_{2} = T_{1} {(\frac{V_{1}}{V_{2}})}^{γ - 1} = 300 {(16)}^{\frac{7}{5} - 1} = T_{2} = 300 {(2^{4})}^{2 / 5} = T_{2} = 300 \times 2^{4 \times \frac{2}{5}} \end{array}$

During isobaric process

$\begin{array}{l} V = \frac{n R T}{P} \\ \Rightarrow V \propto T \\ \Rightarrow \frac{V_{2}}{V_{3}} = \frac{T_{2}}{T_{3}} \\ \Rightarrow \frac{V_{2}}{2 V_{2}} = \frac{300 \times 2^{4 \times \frac{2}{5}}}{T_{3}} \\ \Rightarrow T_{3} = 2 \times 300 \times 2^{8 / 5} = 1818.85 = 1819 K \end{array}$