Question

A glass bulb of volume 400${cm}^3$ is connected to another bulb of volume 200${cm}^3$ by means of a tube of negligible volume. The bulbs contain dry air and are both at a common temperature and pressure of ${20}^0$C and 1.00 atm. The larger bulb is immersed in steam at ${100}^0$C; the smaller, in melting ice at $0^0$. Find the final common pressure.

• A. 1.563 atm
• B. 2.05 atm
• C. 2.36 atm
• D. 1.134 atm

Answer 1

The correct option is D 1.134 atm

$P_i=1atm$

$T_i={20}^{o}C=293K$

$V_i=0.6L$

$\therefore \frac{PiVi}{T_i}=\frac{0.6}{293}$

let final press be P

$T_A=373K$

$T_B=273K$

$\therefore \frac{P{V}_A}{T_A}+\frac{P{V}_B}{T_B}=\frac{P_i{V}_i}{T_i}$

$\Rightarrow P(\frac{0.4}{373}+\frac{0.2}{273})=\frac{0.6}{293}$

$\Rightarrow P(\frac{2}{373}+\frac{1}{273})=\frac{3}{293}$

$\Rightarrow P=\frac{3\times 373\times 273}{293\times 919}$

$\approx 1.134atm$