# 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?

Jul 22, 2018

The volume is $= \frac{1}{30} m L$

#### Explanation:

Apply Boyle's Law

$\text{Pressure "xx" Volume "=" Constant}$

The temperature being constant

${P}_{1} {V}_{1} = {P}_{2} {V}_{2}$

The initial volume is ${V}_{1} = 2 L$

The initial pressure ${P}_{1} = 1 a t m$

The final pressure is ${P}_{2} = 6 \cdot {10}^{4} a t m$

The final volume is

${V}_{2} = \frac{1}{6 \cdot {10}^{4}} \cdot 2 = \frac{1}{3} \cdot {10}^{-} 4 L = \frac{1}{30} m L$

Jul 22, 2018

Rather small, a fraction of a millilitre….

#### Explanation:

Old Boyle's Law insists that ${P}_{1} {V}_{1} = {P}_{2} {V}_{2}$...

And so …………….

${V}_{2} = \frac{{P}_{1} {V}_{1}}{P} _ 2 = \frac{2.00 \cdot L \times 1.00 \cdot a t m}{6.00 \times {10}^{4} \cdot a t m} = 3.33 \times {10}^{-} 5 \cdot L \cong 0.03 \cdot c {m}^{3}$

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

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

$P \setminus \propto \frac{1}{V}$

PV=\text{constant

${P}_{1} {V}_{1} = {P}_{2} {V}_{2}$

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

$1 \setminus \times 2 = 6 \setminus \times {10}^{4} \setminus \times {V}_{2}$

${V}_{2} = \frac{1}{3} \setminus \times {10}^{- 4}$

$= 3.33 \setminus \times {10}^{- 5} \setminus \setminus \textrm{< r s}$

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