If a gas at occupies 2.60 liters at a pressure of 1.00 atm, what will be its volume at a pressure of 3.50 atm?

Question

Wyatt Adkins · Accepted Answer

Approx. #3/4*L.# #"Boyle's Law"# gives us the relationship #Pprop1/V# for constant #T#, and for constant amount of gas. Thus #PV=k#. For constant temperature, and constant amount of gas, as pertains in this scenario, we write: #P_1V_1=P_2V_2# And thus #V_2=(P_1V_1)/(P_2)=(2.60*Lxx1.00*atm)/(3.50*atm)=??L#. The best thing about using these simple experimental laws of proportionality is that we can persist with using intuitively easy units such as #"litres"# and #"atmospheres"#. It is intuitively obvious that the volume of a gas sample under compression should reduce.

Natalie Anton · Answer

undefined Using Boyle's Law, $P_1V_1 = P_2V_2$, where $P_1$ and $V_1$ are the initial pressure and volume, and $P_2$ and $V_2$ are the final pressure and volume, respectively, we can solve for $V_2$:

$P_1V_1 = P_2V_2$

$1.00 \, \text{atm} \times 2.60 \, \text{L} = 3.50 \, \text{atm} \times V_2$

$V_2 = \frac{1.00 \, \text{atm} \times 2.60 \, \text{L}}{3.50 \, \text{atm}}$

$V_2 = 0.743 \, \text{L}$

Therefore, the volume of the gas at a pressure of 3.50 atm would be approximately 0.743 liters.