Given that the volume of cylinder at STP = 11.2 L
Standard temperature = 273 K
Standard pressure = 1 atm
Calculating the volume of gas at 202.6 kPa and 300. K:
Applying the combined gas law,
[tex]\frac{P_{1}V_{1}}{T_{1}} =\frac{P_{2}V_{2}}{T_{2}}[/tex]
[tex]P_{1} =1 atm[/tex]
[tex]V_{1}=11.2 L[/tex]
[tex]T_{1}=273K[/tex]
[tex]P_{2}[/tex]= [tex]202.6 kPa*\frac{1 atm}{101.325 kPa} =2 atm[/tex]
[tex]V_{2}= ? [/tex]
[tex]T_{2}=300. K[/tex]
Plugging in the values and calculating final volume:
[tex]\frac{(1atm)(11.2L)}{(273 K)} =\frac{(2atm)(V_{2})}{(300K)}[/tex]
[tex]V_{2}=6.15 L[/tex]
Therefore the correct answer will be, b) 6.15 L
