Question

What is the network done (ln calories) by $1$ mole of monoatomic ideal gas in a process described by $1,2,3,4$ in given $V-T$ graph?
Use : $R=$ $2$ cal/mole K
ln $2$ = $0.7$

• A. $+630$ cal
• B. $-660$ cal
• C. $+660$ cal
• D. $+600$ cal

Answer 1

The correct option is A $+630$ cal

for process one $V_1=10L{V}_2=20LT=300K$

$W=-nRTln\frac{V_2}{V_1}$

$=-1\times 2\times 300\times ln\frac{20}{10}$

$=-600\times ln2$

$=-600\times 0.7$

$=-420cal$

for process two (assuming adiabatic process) process is expansion work $-ve$

$w\because △D$

$\Rightarrow w=nc(7_2-7_1)ln\left(\frac{V_2}{V_1}\right)$

$=1\times \frac{3}{2}a(600-300)ln\left(\frac{40}{20}\right)$

$=1\times \frac{3}{2}\times 2(600-300)\times ln2$

$=900cal\times 0.7=630$

$\Rightarrow w=-900cal=-630cal$

for process thrice $V_1=40L{V}_2=10LT=600k$

$w=-nRTln\frac{V_2}{V_1}$

$=-1\times 2\times 600ln\frac{10}{40}$

$=-1\times 1200\times (-ln{2}^2)$

$=+1200\times 2ln2=1680cal$

for process four $w=0$ (volume always e$=0$)

net work $=-420+1680-630=+630cal$

Please check your options, the ans obtained is not mentioned in the question.