(D) The ideal gas equation is $PV = nRT$. For $n = 1 \ mol$,the equation becomes $P = (RT) \times (V^{-1})$. Comparing this with the equation of a str

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(D) The ideal gas equation is $PV = nRT$. For $n = 1 \ mol$,the equation becomes $P = (RT) \times (V^{-1})$. Comparing this with the equation of a straight line $y = mx + c$,where $y = P$ and $x = V^{-1}$,we get the slope $m = RT$. Given that the slope $m = 32.8 \ L \ atm \ mol^{-1}$ and $R = 0.082 \ L \ atm \ K^{-1} \ mol^{-1}$. Therefore,$RT = 32.8 \ L \ atm \ mol^{-1}$. $T = \frac{32.8 \ L \ atm \ mol^{-1}}{0.082 \ L \ atm \ K^{-1} \ mol^{-1}} = 400 \ K$.

Class 11 Chemistry States of Matter Ideal gas equation and Related gas laws

Medium

At $T \ (K)$ for one mole of an ideal gas,the graph of $P$ (on $y$-axis) and $V^{-1}$ (on $x$-axis) gave a straight line with a slope of $32.8 \ L \ atm \ mol^{-1}$. What is the temperature (in $K$)? $\left(R=0.082 \ L \ atm \ K^{-1} \ mol^{-1}\right)$

AP EAMCET 2024 All AP EAMCET

A
$600$
B
$200$
C
$800$
D
$400$

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Ideal gas equation and Related gas laws

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At $T \ (K)$ for one mole of an ideal gas,the graph of $P$ (on $y$-axis) and $V^{-1}$ (on $x$-axis) gave a straight line with a slope of $32.8 \ L \ atm \ mol^{-1}$. What is the temperature (in $K$)? $\left(R=0.082 \ L \ atm \ K^{-1} \ mol^{-1}\right)$

Similar Questions

Which of the following laws is followed by a hot air balloon used for sports and meteorology?

One litre of a gas weighs $2 \ g$ at $300 \ K$ and $1 \ atm$ pressure. If the pressure is made $0.75 \ atm$,at which of the following temperatures will one litre of the same gas weigh $1 \ g$?

Three flasks of equal volumes contain $CH_4$,$CO_2$,and $Cl_2$ gases respectively. They will contain equal number of molecules if

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