Question

An electric bulb of volume $250{\text{cm}}^3$ was sealed off during manufacture at a pressure of ${10}^{-3}\text{mm}$ of mercury at ${27}^{\circ }C$. Compute the number of air molecules contained in the bulb. Given that $R=8.31\text{J/mole-K}$ and $N_A=6.02\times {10}^{23}$ per mole

(correct answer + 2, wrong answer - 0.50)

• A. $8\times {10}^{15}$
• B. $7\times {10}^{15}$
• C. $6.92\times {10}^{15}$
• D. $9.87\times {10}^{15}$

Answer 1

The correct option is A $8\times {10}^{15}$

Let there be $n$ moles of air in the bulb.
Hence, from ideal gas equation $PV=nRT$
Pressure $\left(P\right)=\frac{{10}^{-3}}{760}\times 1.01\times {10}^{5}Pa$
Volume $\left(V\right)=250\times {10}^{-6}{m}^3$
Universal gas constant $\left(R\right)=8.31\text{J/mol - K}$
Temperature $\left(T\right)={27}^{\circ }C=300K$
$\frac{{10}^{-3}}{760}\times 1.01\times {10}^5\times (250\times {10}^{-6})=n\left(8.31\right)\left(300\right)$
$n=1.33\times {10}^{-8}=\frac{N}{N_A}=\frac{N}{6.023\times {10}^{23}}$
$N\approx 8\times {10}^{15}$