One mole of an ideal gas is subjected to a process in which $P = \frac{1}{8.21}$ $V$ 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.

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Solution

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

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As, PV=nRT⇒PV=RT [∵ n=1 ]Given, P(8.21) = V⇒P(8.21)=RTP⇒P2(8.21)=(0.0821)T⇒T=100P2At 10 atm T will be highest, T = (100)(100) = 104 K

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