Question

Three different processes that can occur in an ideal monatomic gas are shown in the P vs V diagram. The paths are labelled as A → B, A → C and A → D. The change in internal energies during these processes are taken as$E_{AB},E_{AC}and{E}_{AD}$ and the work done as $W_{AB},W_{AC}and{W}_{AD}$. The correct relation between these parameters are:

• A. $E_{AB}=E_{AC}=E_{AD},W_{AB}>0,W_{AC}=0,W_{AD}<0$
• B. $E_{AB}>E_{AC}>E_{AD},W_{AB}<W_{AC}<W_{AD}$
• C. $E_{AB}<E_{AC}<E_{AD},W_{AB}>0,W_{AC}>W_{AD}$
• D. $E_{AB}=E_{AC}=E_{AD}{,W}_{AB}>0,W_{AC}=0,W_{AD}>0$

Answer 1

The correct option is A

$E_{AB}=E_{AC}=E_{AD},W_{AB}>0,W_{AC}=0,W_{AD}<0$

Step 1: Draw the diagram:

1. The only contribution to energy for a monatomic ideal gas (such as helium, neon, or argon) is translational kinetic energy.
2. The average translational kinetic energy of a single atom is provided by the equation $K_{avg}=\frac{3}{2}kT$, and depends only on the gas temperature.

Therefore,$E_{AB}=E_{AC}=E_{AD}$

Step 2: Apply the formula

$\Delta U=\frac{nfR}{2}(T_f–T_i)$, Where, $n=$number of moles, $T_f\&T_i=$Final and initial temperature, $R=$Gas constant, and $U=$Heat

$W=P∆V$

So,

1. $AB\to$volume is increasing $w_{AB}>0$
2. $AC\to$volume is constant $w_{AC}=0$
3. $AD\to$volume is decreasing $w_{AD}<0$
   $W_{AB}>0,W_{AC}=0,W_{AD}<0$

Hence, the correct option is (A).