"This\\space problem\\space is\\space solved\\space using\\space the\\space Ideal\\space Gas\\space Law\\space formula.\\newline\n\\space \\space\\space PV=nRT\\newline n\\space is\\space number\\space of\\space moles\\space which\\space is=\\dfrac{21g}{28.02}"
"=0.75mol"
"Convert\\space volume(164ml)\\space into\\space litres(L)\\newline\\space \\space \\space \\dfrac{164}{1000}=0.164L\\newline Rearraging\\space the\\space formula\\space we \\space get,\\newline\\space \\space \\space T=\\dfrac{PV}{nR} \\newline \\newline \\space \\space \\space T=\\dfrac{6*0.164}{0.75*0.082}=16\\newline The\\space absolute\\space temperature\\space is\\space16K"
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Calculate the quantities for ideal gas given below. (R=0.082 L atm mol-1 K-1) The absolute temperature of 21 g Nitrogen gas at which gas occupies 164 mL at 6 atm ii. The pressure in atm, if 8.25 X 10 -2 mole gas occupies a volume of 255 mL at 115ºC. iii. The molecular weight of 12.3 g gas if 5.49 L at 35ºC have pressure 11.25 atm.
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(R=0.082 L atm mol-1 K-1) Calculate the pressure in atm, if 8.25 X 10 -2 mole gas occupies a volume of 255 mL at 115ºC.
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