Question

14g oxygen at $0^{\circ }C$ and 10 atm are subjected to reversible adiabatic expansion to a pressure of 1 atm. What is the work done?

• A. -12.80 liter atm
• B. -11.82 liter atm
• C. -11.90 liter atm
• D. -12 liter atm

Answer 1

The correct option is B -11.82 liter atm

$P_1$ = 10 atm at $T=273k$ for $\frac{14}{32}$ mole of $O_2$

$P_2$ = 1 atm at $T=T_{2}k$ for $\frac{14}{32}$ mole of $O_2$

For adiabatic expansion we have $T^{\gamma }{P}^{1-\gamma }$ = constant

$\therefore {\left(\frac{T_1}{T_2}\right)}^{\gamma }={\left(\frac{P_2}{P_1}\right)}^{1-\gamma }$

or $\gamma =\left(\frac{T_1}{T_2}\right)=(1-\gamma )log\left(\frac{P_2}{P_1}\right)$

or $1.4log\frac{273}{T_2}=(1-1.4)log\left(\frac{1}{10}\right)$

$\therefore T_2=141.4k$

Work done in adiabatic expansion = $\frac{nR}{(\gamma -1)}(T_2-T_1)$

= $\frac{14}{32}\times \frac{0.0821(141.4-273)}{(1.4-1)}$

$W_{rev}=-11.82literatm$