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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| $P \ (bar)$ | $\frac{PV}{RT}$ |
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