Question

One mole of an ideal monatomic gas has initial temperature $T_0$ is made to go through the cycle $abca$ as shown in the given figure. If $U$ denotes the internal energy, then choose the correct alternative.

• A. $U_c>U_b>U_a$
• B. $U_c-U_b=3R{T}_0$
• C. $U_c-U_a=\frac{9R{T}_0}{2}$
• D. $U_b-U_a=\frac{3R{T}_0}{2}$

Answer 1

Answer: (A), (B), (C), (D)

$U_b-U_a=\frac{3R{T}_0}{2}$

Internal energy is a state function.(Doesnot depend on path)
We know, $\Delta U=n{C}_v\Delta T$
From $a$ to $b$, pressure is contant
$\Rightarrow V\propto T$
Volume changes from $V_o$ to $2V_o$
$\Rightarrow$ temperature changes from $T_o$ to $2T_o$ .
For a monoatomic gas, $C_v=\frac{3}{2}R$
$U_b-U_a=n{C}_v\Delta T$
$=1\times \frac{3}{2}R(2T_o-T_o)=\frac{3}{2}R{T}_o$....(i)
Option (d) is correct

From b to c, volume is constant.
$\Rightarrow P\propto T$
Pressure changes from $P_o$ to $2P_o$
$\Rightarrow$ temperature changes from $2T_o$ to $4T_o$.
$U_c-U_b=n{C}_v\Delta T$
$=1\times \frac{3}{2}R(4T_o-2T_o)=3R{T}_o$....(ii)
Option (b) is correct
Also option (a) is correct

For c to a, we add equations (i) and (ii)
$U_c-U_a=(U_c-U_b)+(U_b-U_a)$
$=3R{T}_o+\frac{3}{2}R{T}_o=\frac{9R{T}_0}{2}$
Option (c) is also correct.