Question

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?

Answer 1

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`

Answer 2

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

`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`