Question

What is the pressure in millimeters of mercury of 0.0130 molmol of helium gas with a volume of 210. mLmL at 55 ∘C∘C? (Hint: You must convert each quantity into the correct units (LL, atmatm, molmol, and KK) before substituting into the ideal gas law.)

Answer 1

Answer : The pressure of the helium gas is, 1269.2 mmHg

Explanation :

To calculate the pressure of the gas we are using ideal gas equation:

$PV=nRT$

where,

P = Pressure of $He$ gas = ?

V = Volume of $He$ gas = 210. mL = 0.210 L    (1 L = 1000 mL)

n = number of moles $He$ = 0.0130 mole

R = Gas constant = $0.0821L.atm/mol.K$

T = Temperature of $He$ gas = $55^oC=273+55=328K$

Putting values in above equation, we get:

$P\times 0.210L=0.0130mole\times (0.0821L.atm/mol.K)\times 328K$

$P=1.67atm=1269.2mmHg$

Conversion used : (1 atm = 760 mmHg)

Thus, the pressure of the helium gas is, 1269.2 mmHg