Question

If $Pdv/sP$(where P denotes pressure in atm and d denotes density in gm/L) is plotted for He gas (assume ideal) at a particular temperature. If ${{\left[ \dfrac{d}{dP}\left( Pd \right) \right]}_{P=8.21atm}}=5$ then the temperature will be:
A) 160 K
B) 320 K
C) 80 K
D) None of these.

Answer 1

Hint: In order to solve this question we have to use a density version of the ideal gas law, after writing formula we will differentiate the formula after rearranging the formula to match with formula that is given in the question.
Formula used:
$PM=dRT$

Complete answer:
We know that formula for the ideal gas law equation.
$PM=dRT$
Where, P = pressure (in atm)
T = temperature
R = ideal gas law constant
Value of R is $\left( 0.0821\dfrac{atm}{mol} \right)$
M = molar mass of the substance
d = density
Now let’s rearrange the given formula to convert in the form as given in the question.
$d=\dfrac{PM}{RT}$
Now in order to rearrange as given in the question, let’s multiply both side with the pressure (P)
$Pd=\dfrac{M{{P}^{2}}}{RT}$
Now let’s differentiate with respect to the pressure P.
$\begin{align}
  & \dfrac{d}{dP}\left( Pd \right)=\dfrac{M}{RT}\dfrac{d}{dP}\left( {{P}^{2}} \right) \\
 & \dfrac{d}{dP}\left( Pd \right)=\dfrac{2MP}{RT}...\left( 1 \right) \\
\end{align}$
Now at the pressure P = 8.21 atm, the value of $\dfrac{d}{dP}\left( Pd \right)$ is given as, 5.
And the value of R is $\left( 0.0821\dfrac{atm}{mol} \right)$and the molar mass of the helium is $M=\dfrac{4g}{mol}$
Now let’s substitute all the values in the equation (1)
$\begin{align}
  & \Rightarrow 5=\dfrac{2\times 4\times 8.21}{0.0821\times T} \\
 & \Rightarrow T=\dfrac{8\times 8.21}{0.0821\times 5} \\
 & \therefore T=160K \\
\end{align}$

Therefore the correct option is (A) 160K .

Note:
In order to solve this question first we use the formula for the ideal gas law which is $PM=dRT$after that we rearrange formula as $Pd=\dfrac{M{{P}^{2}}}{RT}$ then after we differentiate this equation with respect to the pressure P and get a new formula as,
$\dfrac{d}{dP}\left( Pd \right)=\dfrac{2MP}{RT}$ and by putting value in the above equation we get the temperature (T) AS 160 Kelvin.