Question

Calculate the final temperature of a sample of argon of mass 12.0 g that is expanded reversibly and adiabatically
from 101 at 273.15 K to 3.0L​

Answer 1

Answer:

For reversible adiabatic process $TV^{r-1} =constant$

where $r=\frac{n+2}{n}$

n=degree of freedom

argon is a mono atomic gas so the degree of freedom is 3

$r=\frac{3+2}{3} =\frac{5}{3}$

Here $T_{1} =273.15$

$T_{2} =?$

$V_{1}=1.0dm^{3}=0.001m^{3}$

$V_{2} =3dm^{2} =0.003m^{3}$

$T_{1}V^{r-1} _{1}  =273.15*(0.001)^{\frac{5}{3} -1}$

                                =273.15*0.01km ^2

                                =2.7315K.m^2

$T_{2} V^{r-1} _{2} =T_{2} *(0.003m^3)^{\frac{5}{3}-1 }$

                                  $=T_{2} *0.020m^{2} .k$

$T_{1} V_{1} ^{r-1} =T_{2} V_{2} ^{r-1}$

2.7315km^2=T2*0.020km^2

$T_{2} =136.57K$