Synthetic diamonds can be manufactured at pressures of #6.00 times 10^4# atm. If we took 2.00 liters of gas at 1.00 atm and compressed it to a pressure #6.00 times 10^4# atm, what would the volume of that gas be?

3 Answers
Jul 22, 2018

Answer:

The volume is #=1/30mL#

Explanation:

Apply Boyle's Law

#"Pressure "xx" Volume "=" Constant"#

The temperature being constant

#P_1V_1=P_2V_2#

The initial volume is #V_1=2L#

The initial pressure #P_1=1atm#

The final pressure is #P_2=6*10^4atm#

The final volume is

#V_2=1/(6*10^4)*2=1/3*10^-4L=1/30mL#

Jul 22, 2018

Answer:

Rather small, a fraction of a millilitre….

Explanation:

Old Boyle's Law insists that #P_1V_1=P_2V_2#...

And so …………….

#V_2=(P_1V_1)/P_2=(2.00*Lxx1.00*atm)/(6.00xx10^4*atm)=3.33xx10^-5*L~=0.03*cm^3#

And note our assumptions...we ASSUME (perhaps wrongly) that the gas would not undergo a phase change.

Answer:

#3.33\times 10^{-5}\ \text{ltrs#

Explanation:

Assuming gas is ideal undergoing a constant temperature process then

According to Boyle's Law, at constant temperature #T#, the pressure #P# of a constant mass of an ideal gas is inversely proportional to its volume #V#. Mathematically given as

#P\prop 1/V#

#PV=\text{constant#

#P_1V_1=P_2V_2#

As per given data, initial pressure #P_1=1\ text{atm# , initial volume #V_1=2\ text{ltrs# & final pressure #P_2=6\times 10^4\ text{atm#. Substituting these values in above equation, we get

#1\times 2=6\times 10^4\times V_2#

#V_2=1/3\times 10^{-4}#

#=3.33\times 10^{-5}\ \text{ltrs}#

The final volume of gas becomes #V_2=3.33\times 10^{-5}\ \text{ltrs#