Question

A glass bulb of volume $400c{m}^3$ is connected to another of volume $200c{m}^3$ by means of a tube of negligible volume. The bulbs contain dry air and both are at a common temperature and pressure of ${20}^{\circ }C$ and $1.00$ $atm$ respectively. The larger bulb is immersed in steam at ${10}^{\circ }C$ and the smaller in melting ice at $0^{\circ }C$. Find the final common pressure.

• A. $6.13atm$
• B. $7.13atm$
• C. $2.13atm$
• D. $1.13atm$

Answer 1

The correct option is D $1.13atm$

Since the total number of moles must remain constant,

From $n=\frac{PV}{RT}$,

$\frac{1\left(400\right)}{293}+\frac{1\left(200\right)}{293}=P(\frac{400}{283}+\frac{200}{273})$

$P=\frac{\frac{3}{293}}{\frac{2}{283}+\frac{1}{273}}=1.13atm$

Answer is $1.13atm$