Asked by Anonymous
What pressure (in atmospheres) will be exerted by 1.3 moles of gas in a 13 litre flask at 22 degrees Celcius? (Gas constant (R) = 0.0821 L atm K^-1 mol^-1)
Answers
Answered by
MathMate
Use ideal gas law:
PV=nRT (T in °K)
=>
P=nRT/V
=1.3 mol *0.0821 L atm °K<sup>-1</sup> mol<sup>-1</sup> *(22+273.16)°K/13 L
=2.423 atm
=2.4 atm (approx.)
PV=nRT (T in °K)
=>
P=nRT/V
=1.3 mol *0.0821 L atm °K<sup>-1</sup> mol<sup>-1</sup> *(22+273.16)°K/13 L
=2.423 atm
=2.4 atm (approx.)
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