Answer:
[tex]V_2= 1.19L[/tex]
Explanation:
Hello there!
In this case, since the STP conditions are defined by 1 atm (101.3 kPa) and 273 K, it is possible for us to use the combined gas law for this problem as we are given variable pressure, temperature and volume:
[tex]\frac{P_1V_1}{T_1} =\frac{P_2V_2}{T_2}[/tex]
In such a way, solving for V2 as the final volume, we obtain:
[tex]V_2=\frac{P_1V_1T_2}{P_2T_1} =\frac{156kPa*2.45L*273.15K}{101.3kPa*(594+273)K}\\\\V_2= 1.19L[/tex]
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