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One mole of an ideal gas is subjected to a process in which $$P = \dfrac{1}{8.21}$$ $$V$$ where $$P$$ is in atm and a $$ V$$ in litre. If the process is operating from $$1$$ atm to finally $$10$$ atm (no higher pressure achieved during the process) then that would be the maximum temperature obtained & at what instant will it occur in the process.


Solution

$$ \text{As,     PV=nRT}   \\  \Rightarrow \text{PV=RT     [$\because$ n=1 ]} \\  \text{Given,  P(8.21) = V} \Rightarrow P (8.21) = \cfrac{RT}{P} \Rightarrow P^{2}(8.21) = (0.0821)T \\ \Rightarrow T= 100 P^{2} \\ \text{At 10 atm T will be highest,     T = (100)(100) = 10$^{4}$ K}$$

Chemistry

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