Answer : The pressure of helium gas is, 926.4 mmHg
Explanation :
To calculate the pressure of helium gas we are using ideal gas equation:
[tex]PV=nRT\\\\PV=\frac{w}{M}RT[/tex]
where,
P = pressure of helium gas = ?
V = volume of helium gas = 645 mL = 0.645 L
T = temperature of helium gas = [tex]31^oC=273+31=304K[/tex]
R = gas constant = 0.0821 L.atm/mole.K
w = mass of helium gas = 0.126 g
M = molar mass of helium gas = 4 g/mole
Now put all the given values in the ideal gas equation, we get:
[tex]P\times 0.645L=\frac{0.126g}{4g/mole}\times (0.0821L.atm/mole.K)\times (304K)[/tex]
[tex]P=1.2189atm=926.4mmHg[/tex]
Conversion used : (1 atm = 760 mmHg)
Therefore, the pressure of helium gas is, 926.4 mmHg
Given:
= 273+31
= 0.645 L
By using the Ideal gas equation, we get
→ [tex]PV = nRT[/tex]
or,
→ [tex]PV = \frac{w}{M}RT[/tex]
By substituting the values, we get
→ [tex]P\times 0.645=\frac{0.126}{4 }\times (0.0821)\times (304 )[/tex]
→ [tex]P = 1.2189 \ atm[/tex]
[tex]= 926.4 \ mmHg[/tex]
Thus the above response is appropriate.
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