A 20 L container holds 0.650 mol of helium at 37 degree Celsius at a pressure of 628.3 bar find the new pressure inside the container if the volume is reduced to 12Lyemperature increased to 177 degree celsius and 1.25 mol of additional helium is added to it
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Answer: new applied pressure derived from gas equation is 88.84
Explanation:
From gas law equation we have PV=NRT
Where V is volume P is pressure N is no of moles T is temp in Kelvin
R is constant.
As P,V, T and n are changing
Hence we have to frame Relationship between them in terms of pressure of P1 and P2 for the initial and final pressure
As, P= nT/V
MOLAR MASS AND TEMP. WILL BE DIRECTLY PROPORTIONAL TO Pressure and indirectly proportional to volume.
AS V1 and V2 are 20 and 12 litres
And T1 and T2 are 37 + 273 = 310K and 177+ 273 = 450k
n1 and n2 are 0.65 mol and 1.25 mol+ 0.65mol = 1.90 mol
P1 is 628.3 bar. P2 will be new pressure.
P1/P2 = n1T1V2/n2T2V1
P2=628.3*0.141
= 88.84bar
HENCE pressure 88.84
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