Asample of a gas in a rigid cylinder with a movable piston has a volume of 11.2 liters at stp. what is the volume of this gas at 202.6 kpa and 300. k? a) 5.10 l b) 6.15 l c)22.4 l d) 24.6 l

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,

=  

Plugging in the values and calculating final volume:

Therefore the correct answer will be, b) 6.15 L

The the volume of this gas at 202.6 kPa and 300 K is 6.15 L.

The combined gas law,

Where,

P1 = 1atm             P2 = 202.6 kPa = 2atm

V1 = 11.2 L            V2 = ?

T1 = 273 K            T2 = 300 K

Put the value in formula we get,

Hence, we can conclude that, The the volume of this gas at 202.6 kPa and 300 K is 6.15 L.

To know more about gas law, refer to the link:

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