.54 mol of H2 is contained in a 2.00 L container at 20.0 oC. What is the pressure in the container in atm?

.54 mol of H2 is contained in a 2.00 L container at 20.0 oC. What is the pressure in the container in atm?

2 Answers
Mar 4, 2018

6.5 atm6.5atm

Explanation:

Using ideal gas law to calculate the pressure of the gas,

So,PV=nRTPV=nRT

Given values are,V=2L,n=0.54 mole,T=(273+20)=293KV=2L,n=0.54mo,T=(273+20)=293K

Using,R=0.0821 LR=0.0821L atm mol^-1K^-1

We get,P=6.5 atmP=6.5atm

Mar 4, 2018

P_"container"=6.50*atmPcontainer=6.50atm

Explanation:

We assume Ideality....and thus...P=(nRT)/VP=nRTV

=(0.54*molxx0.0821*(L*atm)/(K*mol)xx293.15*K)/(2.00*L)=0.54mol×0.0821LatmKmol×293.15K2.00L

Clearly, I used .........

"absolute temperature"="degree Celsius + 273.15"*Kabsolute temperature=degree Celsius + 273.15K

And the expression gives us an answer with units of pressure, as is required....

=(0.54*cancel(mol)xx0.0821*(cancelL*atm)/cancel(K*mol)xx293.15*cancelK)/(2.00*cancelL)