Question

The adiabatic compression ratio in a carnot reversible cycle is $9$ when temperature of the source is ${227}^{\circ }C.$ The temperature of the sink is (given $\gamma =1.5)$

• A. ${166.67}^{\circ }C$
• B. $-{166.67}^{\circ }C$
• C. ${106.33}^{\circ }C$
• D. $-{106.33}^{\circ }C$

Answer 1

The correct option is D $-{106.33}^{\circ }C$

Here, it is given the compression ratio i.e. $\frac{V_2}{V_1}=9$
Also $T_1=227+273=500\text{K},\gamma =1.5$
We have to find $T_2$
For an adiabatic process $T_2{V}_2^{\gamma -1}=T_1{V}_1^{\gamma -1}$
$\Rightarrow T_2=T_1{\left(\frac{V_1}{V_2}\right)}^{\gamma -1}$
$T_2=T_1{\left(\frac{1}{9}\right)}^{1.5-1}$
$\Rightarrow T_2=500\times \frac{1}{3}$
$T_2=166.67\text{K}$
$T_2=166.67-273$
$T_2=-{106.33}^{\circ }C$