An electric blub of volume 250 cm3 was sealed off during manufacture at a pressure of 10-3 mm of mercury at 27-C- Compute the number of air molecules contained in the bulb- Given that R - 8-31J-mol-K and NA - 6-02 - 1023 per mol-

Let there be n moles of air in the bulb. #160;Hence from PV=nRT 10 ^ 3 760 X1.01X 10 ^ 5 X(250X 10 ^ 6 )=n(8.31)(300) n=1.333X 10 ^ 8 = N ...

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Standard X

Physics

Ideal Gas Equation

An electric b...

Question

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

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Solution

Let there be n moles of air in the bulb.

Hence from $P V = n R T$

$\frac{10^{- 3}}{760} X 1.01 X 10^{5} X (250 X 10^{- 6}) = n (8.31) (300)$

$n = 1.333 X 10^{- 8} = \frac{N}{N_{A}} = \frac{N}{6.023 X 10^{23}}$

$N \approx 8 X 10^{15}$

Answer is $8 X 10^{15}$

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An electric blub of volume 250 cm3 was sealed off during manufacture at a pressure of 10−3 mm of mercury at 27∘C. Compute the number of air molecules contained in the bulb. Given that R = 8.31J/mol-K and NA=6.02×1023 per mol.

Let there be n moles of air in the bulb. Hence from PV=nRT10−3760X1.01X105X(250X10−6)=n(8.31)(300)n=1.333X10−8=NNA=N6.023X1023N≈8X1015Answer is 8X1015

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

Let there be n moles of air in the bulb.

Hence from $P V = n R T$

$\frac{10^{- 3}}{760} X 1.01 X 10^{5} X (250 X 10^{- 6}) = n (8.31) (300)$

$n = 1.333 X 10^{- 8} = \frac{N}{N_{A}} = \frac{N}{6.023 X 10^{23}}$

$N \approx 8 X 10^{15}$

Answer is $8 X 10^{15}$