Question

One mole of an ideal monatomic gas has initial temperature $T_0$, is made to go through the cycle $abca$ as shown in fig. 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)

The correct options are
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 $U_b-U_a=\frac{3R{T}_0}{2}$


Ideal gas equation for one mole of an ideal monatomic gas is
$\frac{PV}{T}=\text{constant}$

For $ab$ process [isobaric]
$\frac{V_a}{T_a}=\frac{V_b}{T_b}\Rightarrow \frac{V_0}{T_0}=\frac{2V_0}{T_b}$
$\Rightarrow T_b=2T_0\Rightarrow T_b>T_a\Rightarrow U_b>U_a$
($\because$ Internal energy $U\propto T$)
$U_b-U_a=C_v\Delta T=\frac{3R}{2}(2T_0-T_0)=\frac{3R{T}_0}{2}$

For process $bc$ (isochoric process)
$\frac{P_b}{T_b}=\frac{P_c}{T_c}\Rightarrow \frac{P_0}{2T_0}=\frac{2P_0}{T_c}\Rightarrow T_c=4T_0$
$\Rightarrow T_c>T_a\Rightarrow U_c>U_a$
$U_c-U_b=\frac{3R}{2}(4T_0-2T_0)=3R{T}_0$

For Process $ca$ :
$U_c-U_a=\frac{3R}{2}(4T_0-T_0)=\frac{9R{T}_0}{2}$
Thus, $a,b,c,d$ are the correct