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At $300 \mathrm{~K}$, the following graph is obtained for one mole of an ideal gas. If its pressure is $10 \mathrm{~atm}$, then its volume (in L) will be
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$0.11$
$\begin{aligned} & \text {} \mathrm{PV}=\mathrm{nRT} \text { or } \mathrm{P}=\frac{1}{\mathrm{~V}} \mathrm{nRT}(\mathrm{y}=\mathrm{mx}+0) \\ & \Rightarrow \text { Slope }=\mathrm{nRT}=1.1 \text { (given) } \\ & \Rightarrow \mathrm{V}=\frac{\mathrm{nRT}}{\mathrm{P}}=\frac{1.1}{10}=0.11 \mathrm{~L}\end{aligned}$
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