Question

two moles of an ideal monoatomic gases are allowed to expand adiabatically and reversibly from 300K and 200 K . The work done in the system is (Cv=12.5 J/K/mole)

Answer 1

Answer : The work done in the system is, -2500 J

Solution :

Formula used for reversible adiabatic process :

$w=n\times C_v\times (T_2-T_1)$

where,

w = work done

$C_v$ = heat capacity = 12.5 J/K mole

$T_1$ = initial temperature = 300 K

$T_2$ = final temperature = 200 K

n = number of moles of gas = 2 moles

Now put all the given values in the above formula, we get the work done in the system.

$w=(2moles)\times (12.5J/Kmole)\times (200-300)K=-2500J$

Therefore, the work done in the system is, -2500 J

Answer 2

"From$\quad the\quad given,\\ Initial\quad temperature\quad =\quad 300\quad K\\ Final\quad temperature\quad =\quad 200\quad K\\ { C }_{ v }\quad =\quad 12.5\quad kj/mole\\ Number\quad of\quad moles\quad of\quad gas\quad n\quad =\quad 2\\ { C }_{ p }\quad =\quad { C }_{ v }\quad -R\\ R\quad =\quad Gas\quad constant$

${ C }_{ p }=\quad 12.5\quad -\quad 8.314\\ \gamma \quad =\quad \frac { { C }_{ p } }{ { C }_{ v } } \quad =\quad \frac { 12.5-8.314 }{ 12.5 } \quad =\quad 1.665\\ \\ Work\quad done\quad W\quad =\quad \frac { nR({ T }_{ 1 }-{ T }_{ 2 }) }{ \gamma -1 } \\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad =\frac { 2\quad \times \quad 8.314\quad \times (300-100) }{ 1.665-1 }$

$\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad =\quad -2.5\quad kJ$"