An electron tube was sealed off during manufacture at a pressure of 1.2× 10-7mm of mercury at 27° C . Its volume is 100 cm3. The number of molecules that remain in the tube is:

Question

An electron tube was sealed off during manufacture at a pressure of 1.2× 10-7mm of mercury at 27° C . Its volume is 100 cm3. The number of molecules that remain in the tube is: (A)  2× 1016  (B)  3× 1015  (C)  3.86× 1011  (D)  5× 1011

Tardigrade Admin · Accepted Answer

PV=nRT ⇒ (1.2× 10-10× 13.6× 103× 9.8)× 10-4 =n× 8.31× (273+27) ∴ number of moles n=6.42× 10-13 Hence, number of molecules =6.42× 10-13× 6.02× 1023 =3.86× 1011

Q. An electron tube was sealed off during manufacture at a pressure of $1.2 \times 10^{- 7} mm$ of mercury at $27^{\circ} C$ . Its volume is 100 cm3. The number of molecules that remain in the tube is:

$2 \times 10^{16}$

$3 \times 10^{15}$

$3.86 \times 10^{11}$

$5 \times 10^{11}$

$2.5 \times 10^{12}$

$P V = n RT$ $\Rightarrow$ $(1.2 \times 10^{- 10} \times 13.6 \times 10^{3} \times 9.8) \times 10^{- 4}$ $= n \times 8.31 \times (273 + 27)$ $∴$ number of moles $n = 6.42 \times 10^{- 13}$ Hence, number of molecules $= 6.42 \times 10^{- 13} \times 6.02 \times 10^{23}$ $= 3.86 \times 10^{11}$

Q. An electron tube was sealed off during manufacture at a pressure of 1.2×10−7mm of mercury at 27∘C . Its volume is 100 cm3. The number of molecules that remain in the tube is:

2×1016

3×1015

3.86×1011

5×1011

2.5×1012

PV=nRT ⇒ (1.2×10−10×13.6×103×9.8)×10−4 =n×8.31×(273+27) ∴ number of moles n=6.42×10−13 Hence, number of molecules =6.42×10−13×6.02×1023 =3.86×1011