Question

The given figure shows the isotherms of a fixed mass of an ideal gas at three temperatures $T_A,T_B$ and $T_C$ then

• A. $T_A>T_B>T_C$
• B. $T_A<T_B<T_C$
• C. $T_B<T_A<T_C$
• D. $T_A=T_B=T_C$

Answer 1

The correct option is B $T_A<T_B<T_C$

From the ideal gas equation,
$PV=nRT$
$\Rightarrow \frac{nRT}{V}=P$
By drawing a straight line parallel to $P$ - axis, value of $\frac{1}{V}$ is fixed.
Thus, $P\propto T$

$\Rightarrow P_C>P_B>P_A$
$\therefore T_C>T_B>T_A$


OR

From the ideal gas equation,
$PV=nRT$
$\Rightarrow \frac{1}{V}=\frac{P}{nRT}$
Comparing with $y=mx$,
$m\propto \frac{1}{T}$
Slope of $A>$ Slope of $B>$ Slope of C
Hence, $T_C>T_B>T_A$