Question

2 moles of an ideal gas A ($C_P$ $=4R$) and 4 moles of an ideal gas B ($C_V$ $=\frac{3R}{2}$) are taken together in a container and allowed to expand reversibly and adiabatically from 49L to 64L, starting from an initial temperature of ${47}^0$C. The final temperature of the gas is :

• A. $4{}^{0}C$
• B. $5{}^{0}C$
• C. $7{}^{0}C$
• D. $8{}^{0}C$

Answer 1

The correct option is C $7{}^{0}C$

${\gamma }_{min}=\frac{n_A{C}_{P_A}+n_B{C}_{P_B}}{n_A{C}_{V_A}+n_B{C}_{V_B}}=\frac{2\times 4R+4\times \frac{5R}{2}}{2\times 3R+4\times \frac{3R}{2}}$$=\frac{18R}{12R}=1.5$

Now,

$T_1{V}_1^{\gamma -1}=T_2{V}_2^{\gamma -1}$
$320\times {\left(49\right)}^{0.5}=T_2\times {\left(64\right)}^{0.5}$

$T_2=\frac{320\times 7}{8}=280K=7^{0}C$