One mole of ideal gas goes through process then change in temperature of gas when volume changes from 1m3 to 2m3 is

As P=2V21+V2V1=1m3 ; V2=2m3 ⇒P1=22=1N/m2 and P2=85N/m2P2V2−P1V1=nRΔTP2V2−P1V1=nRΔT⇒ΔT=85×2−1×1R=115RK

One mole of ideal gas goes through process then change in temperature of gas when volume changes from 1m3 to 2m3 is

As P=2V21+V2V1=1m3 ; V2=2m3 ⇒P1=22=1N/m2 and P2=85N/m2P2V2−P1V1=nRΔTP2V2−P1V1=nRΔT⇒ΔT=85×2−1×1R=115RK

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One mole of ideal gas goes through process ${\text{P}} = \frac{{{\text{2}}{{\text{V}}^{\text{2}}}}}{{{\text{1}} + {{\text{V}}^{\text{2}}}}}$ then change in temperature of gas when volume changes from 1m³ to 2m³ is

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$- \frac{4}{5 R} K$

$\frac{11}{5 R} K$

$\frac{- 5}{2 R} K$

answer is B.

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As $P = \frac{2 V^{2}}{1 + V^{2}}$

V₁=1m³ ; V₂=2m³

$\Rightarrow P_{1} = \frac{2}{2} = 1 N / m^{2} and P_{2} = \frac{8}{5} N / m^{2}$

$\begin{matrix} P_{2} V_{2} - P_{1} V_{1} = n R Δ T \\ P_{2} V_{2} - P_{1} V_{1} = n R Δ T \\ \Rightarrow Δ T = \frac{\frac{8}{5} \times 2 - 1 \times 1}{R} = \frac{11}{5 R} K \end{matrix}$

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One mole of ideal gas goes through process ${\text{P}} = \frac{{{\text{2}}{{\text{V}}^{\text{2}}}}}{{{\text{1}} + {{\text{V}}^{\text{2}}}}}$ then change in temperature of gas when volume changes from 1m³ to 2m³ is