A cylindrical tank has a tight-fitting piston that allows the volume of the tank to be changed. The tank originally contains air with a volume of 0.185 m3 at a pressure of 0.740 atm. The piston is slowly pulled out until the volume of the gas is increased to 0.315 m3. If the temperature remains constant, what is the final value of the pressure?

0.435atm

Explanation:

cylindrical tank has a tight-fitting piston that allows the volume of the tank to be changed. The tank originally contains air with a volume of 0.185 m3 at a pressure of 0.740 atm. The piston is slowly pulled out until the volume of the gas is increased to 0.315 m3. If the temperature remains constant, what is the final value of the pressure?

Given

Initial pressure P1= 0.740atm

Initial volume V1= 0.185 m3

Final pressure P2= ?

Final volume V2= 0.315 m3

At constant temperature, the pressure of a syste is inversely proportional to volume, by Boyles law then

P1V1=P2V2

P2=P1V1/V2

=(0.185*0.740)/0.315

0.1369/0.315

= 0.435atm

Therefore, final pressure is 0.435atm

  1.the balanced   chemical equation is as below

cu +   2 ag(no)₃ → cu(no₃)₂   +2 ag

2.the   much silver   produced   is   50.9 g

step 1: find the   moles of cu

moles =   mass÷   molar mass

  from periodic table the molar mass of cu = 63.5 g/mol

moles is therefore = 15.00 g÷ 63.5 g/mol = 0.236   moles

step 2: use the mole ratio to determine the   moles of ag

from equation above   cu: ag   is 1: 2

therefore the moles of ag   =0.236   moles   x 2/1 =0.472   moles

step 3: find the mass   of ag

mass = moles x molar mass

from periodic table   the molar mass of ag =107.87 g/mol

mass   = 0.472   moles x 107.87 g/mol = 50.9 g

3.   the   much   silver   collected worth is   1.797 oz

  convert   50.9 g   to oz

that is   1 gram = 0.0353 oz

          50.9 grams =?   oz

by cross multiplication

={(50.9 grams x 0.0353 oz) / 1   gram}   =1.797 oz