A sample of neon gas collected at a pressure of 531 mmHg and a temperature of 291 K has a mass of 10.2 grams. What is the volume (L) of the sample?

Question

anonymous · Answer

I'd go with the ideal gas law here:
$$PV=nRT$$You know P, R, and T. You can work out n (the number of moles of the gas).

anonymous · Answer

A more convient form of the ideal gas law in this case: $$PV = {m \over M} RT$$ where m is the mass (in kg) and M is the molar mass (in kg/kmol or g/mol)

anonymous · Answer

i tried the PV=nRT

I got (10.2 mol)(0.0821 l*atm)(291 k)/0.6986 atm

anonymous · Answer

is it 243.089/0.6986

anonymous · Answer

which comes out to 348.82

anonymous · Answer

521 mm-Hg is 69.461 kPa. The molecular mass of Neon is 20.18 kg/kmol. R = 8.314 kJ/kmol-K. Therefore, we can write the ideal gas law as follows. $$69.461 [kPa] *V[m^3] = {10.2 [g] \over 20.18 [g/mol]} 8.314 [{J \over mol-K}] 291 [K]$$ $$V = 0.0176 [m^3]$$ Since, 1 L = 10^-3 m^3. 
$$V = 17.6 [L]$$

anonymous · Answer

I got 17.27

anonymous · Answer

I had to convert mole to mass which is 0.5054