Question

Calculate the total work done along the path 'ABCD' (as shown in figure) for a one mole of monoatomic gas.

• A. $W=-RTln2$
• B. $W=-P_0{V}_0(1+ln2)$
• C. $W=P_0{V}_0(1+ln2)$
• D. $W=-P_0{V}_{0}ln2$

Answer 1

The correct option is A $W=-RTln2$

At A and D the temperatures of the gas will be equal, so
$△E=0,△H=0$
Now $W=W_{AB}+W_{BC}+W_{CD}$

Now,
for process AB is isobaric , so
work= $-P△V$
= $-P_0(2V_0-V_0)$
=$-P_0{V}_0$

along BC, as it is an isothermal process,
$W=$$-RTln\left(\frac{4V_0}{2V_0}\right)$
= $-RTln2$

Along BC volume is doubled and as it is isothermal process so pressure must be halved.
Pressure at C is $0.5\times P_0$
Along CD,
W =$-0.5\times P_0(2V_0-4V_0)$
= $P_0{V}_0$

So,
Total work done ,
$W=W_{AB}+W_{BC}+W_{CD}$
=$-P_0{V}_0$ +(-RT ln2) +( $+P_0{V}_0$ )
= -RT ln2