Question

One mole of an ideal monoatomic gas undergoes a cyclic process 'abca' as shown in figure. If $U$ denotes the internal energy and $W$ the work done by the gas. Then choose the incorrect altenative (s)

• A. $U_c-U_a=10.5R{T}_0$
• B. $U_b-U_a=+4R{T}_0$
• C. $W_{c\to a}=-3P_0{V}_0$
• D. $U_c>U_b>U_a$

Answer 1

Answer: (B), (C)

The correct options are
B $U_b-U_a=+4R{T}_0$
C $W_{c\to a}=-3P_0{V}_0$



(a) $U_c-U_a=n{C}_{v}dT=\left(1\right)\frac{3R}{2}(8T_0-T_0)=\frac{21}{2}R{T}_0$

(b) $U_b-U_a=n{C}_{v}dT$
$=\left(1\right)\frac{3R}{2}(4T_0-T_0)=4.5R{T}_0$

(c) $W_{c\to a}$ cannot be determined, but $\left|W_{c\to a}\right|>\left|W_{a\to b}\right|$

$\left|W_{c\to a}\right|>3P_0{V}_0,$ Hence $\left(C\right)$ is wrong.