Question

2. Use the following combined gas law formula and compute the volume that the butane sample will occupy at STP. (HINT: Convert both temperatures to Kelvin.) Pbutane x Voriginal/Troom = Pstandard x Vfinal/Tstandard

DATA TABLE
Initial mass of butane lighter-21.8 g
Final mass of butane lighter-21.6 g
Volume of butane released-50 mL
Temperature of water-22 oC
Atmospheric pressure-763mm Hg

Answer 1

Answer: STP stands for standard temperature and pressure and refers to conditions of 273K (0 degree centigrade) at 760mmHg.

Explanation: Temperature, T1= 273 + 22= 295K; Pressure, P1= 763mmHg; Volume, V1= 50mL; T2= 273K; P2= 760mmHg; V2= ?

P1*V1/ T1 = P2*V2/ T2

V2= P1*V1*T2/P2*T1

V2= 763*50*273/760*295

∴ V2= 10414950/224200 = 46.45mL

Answer 2

When we put Boyle's law, Charles' law, and Gay-Lussac's law together, we come up with the combined gas law, which shows that: Pressure is inversely proportional to volume, or higher volume equals lower pressure. It is expressed as:

Pbutane x Voriginal/Troom = Pstandard x Vfinal/Tstandard

It seems that the given values is not complete the pressure of butane is not given and also we cannot calculate for it since there is no more than two data about the water.