For $1 \ mol$ of an ideal gas,an isochore is obtained. The slope of the isochore is $0.082 \ atm \ K^{-1}$. What will be its pressure (in $atm$) when the temperature is $12.2 \ K$? $(R = 0.082 \ L \ atm \ mol^{-1} \ K^{-1})$.

  • A
    $10.0$
  • B
    $0.1$
  • C
    $1.0$
  • D
    $0.5$

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