Answer:
initial temperature=[tex]-3.31^{\circ}C[/tex]
Explanation:
Assuming that the given follows the ideal gas nature;
[tex]P_1=0.723atm[/tex]
[tex]V_1=13.5L[/tex]
[tex]T_1=?[/tex]
[tex]P_2=0.612atm[/tex]
[tex]V_2=17.8L[/tex]
[tex]T_2=28 ^{\circ}C =273+28K=301K[/tex]
mole of gass will remain same at any emperature:
[tex]\frac{P_1V_1}{T_1} =\frac{P_2V_2}{T_2}[/tex]
putting all the value we get:
[tex]T_2 =269.7K =-3.31 ^{\circ} C[/tex]
initial temperature=[tex]-3.31^{\circ}C[/tex]
