what volume would 120.0g of nitrogen gas occupy at 59.50 c and 0.566 atm
59.50oC = 59.50 + 273.15 = 332.65 K
"n(N_2)=\\frac{120.0\\ g}{28.01\\ g\/mol}=4.284\\ mol"
"R=0.08206\\frac{L\\cdot{atm}}{mol\\cdot{K}}"
According to the ideal gas law,
"PV=nRT" ;
"V=\\frac{nRT}{P}=\\frac{4.284\\ mol\\times0.08206\\frac{L\\cdot{atm}}{mol\\cdot{K}}\\times332.65\\ K}{0.566\\ atm}=207\\ L"
Answer: 207 L
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