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Sagot :
Answer: [tex]2.49\ m^3[/tex]
Explanation:
Given
Mass of nitrogen present [tex]m=28\ g[/tex]
Temperature [tex]T=27^{\circ}C\equiv 300\ K[/tex]
Pressure [tex]P=785\ mm\ \text{of}\ Hg\ \text{or}\ 1.032\ atm[/tex]
The molar mass of Nitrogen [tex]M=28\ g/mol[/tex]
No of moles of nitrogen present
[tex]n=\dfrac{m}{M}\\\\n=\dfrac{28}{28}\\\\n=1[/tex]
Using [tex]PV=nRT[/tex]
[tex]\Rightarrow 1.032\times V=1\times 8.314\times 300\\\\\Rightarrow V=\dfrac{2494.2}{1.032}\\\\\Rightarrow V=2494.2\ L\ \text{or}\ 2.49\ m^3[/tex]
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