Answer:
Final temperature will be 438.076 K
Explanation:
We have given temperature [tex]T_1=323K[/tex]
Volume [tex]V_1=V\ and\ V_2=\frac{V}{2}[/tex]
As there is no heat transfer so this is an adiabatic process
For and adiabatic process [tex]TV^{\gamma -1}=constant[/tex]
Here [tex]\gamma =1.4[/tex]
So [tex]T_1V_1^{\gamma -1}=T_2V_2^{\gamma -1}[/tex]
[tex]T_2=\left ( \frac{V_1}{V_2} \right )^{\gamma -1}\times T_1[/tex]
[tex]T_2=\left ( \frac{V}{\frac{V}{2}} \right )^{1.4 -1}\times 332=2^{0.4}\times 332=438.076K[/tex]
