Question

For a fixed amount of ideal gas, match the two Lists.

• A. A - Q; B - P; C - R; D - S
• B. A - S; B - R; C - S; D - R
• C. A - Q; B - P; C - S; D - R
• D. A - Q; B - S; C - R; D - Q

Answer 1

The correct option is A A - Q; B - P; C - R; D - S

(A) We know that, $PV=nRT$ ...........(1)
So, at constant $P,V$ is directly proportional to $T$ (Charles' law)
Taking log we get,
$\log V=\log \frac{nR}{P}+\log T$
Comparing the above equation with the equation of a straight line,$Y=C+mx$
As Intercept $\left(\frac{nR}{P}\right)\uparrow ,P\downarrow$
So, the one with less inercept will have higher $P$
Hence, $P_1>P_2>P_3$

(B) Dividing equation (1) by $V^2$ we get,
$\frac{P}{V}=nRT\times \frac{1}{V^2}$
Comparing the above equation with the equation of a straight line,$Y=C+mx$
Slope is $nRT$ and $C=0$
So, higher the $T$, higher the slope.
We can see that curve $T_3$ has the highest slope. So, $T_3$ is the highest.
So, $T_1<T_2<T_3$

(C) Density (d) is given as:
$d=\frac{PM}{RT}$
Comparing the above equation with the equation of a straight line,$Y=C+mx$
Slope is $\frac{M}{RT}$
So, higher the $T$, lower the slope.
So, $T_1$ is the highest as its slope is the lowest.
$T_1>T_2>T_3$

(D) From equation (1) we can write,
$V=\frac{nR\times T}{P}$
Comparing the above equation with the equation of straight line,$Y=C+mx$
Slope $=\frac{nR}{P}$
Higher the slope, smaller the $P$.
$P_3>P_2>P_1$