Question

There are two bulbs filled with an ideal gas. Bulb $1$ has a volume of $5\text{L}$ and a pressure of $9$ atm, bulb $2$ has a volume of $10\text{L}$ and a pressure of $6$ atm. When these bulbs are connected, what is the final pressure if the temperature remains constant?

• A. 15 atm
• B. 7 atm
• C. 12 atm
• D. 21 atm

Answer 1

The correct option is B 7 atm

Now as the question says temperature is constant which let's us to use Boyle's law
For Bulb 1:
Product of pressure and volume $(PV{)}_1$ = ($5L$)($9atm$)
=$45atmL$

For Bulb 2:
Product of pressure and volume $(PV{)}_2$= ($10L$)($6atm$)
=$60atmL$

Now when bulbs are connected the total volume will be $5L+10L$ =$15L$ = $V_3$ and the new pressure will be obtained by summing up the products of pressure and volume of two bulbs.
i.e. $(PV{)}_3$ = $(PV{)}_1$+ $(PV{)}_2$
$P_3=\frac{5\times 9+10\times 6}{15}$ $atm$
$=7atm$