In a certain region of space there are only 5 molecules per cm $^{3}$ on an average. The temperature is 3K. What is the average pressure of this very dilute gas?

2.07 \times 10^{- 16} P a

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1.07 \times 10^{- 16} P a

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3.07 \times 10^{- 16} P a

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4.07 \times 10^{- 16} P a

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Solution

The correct option is D $2.07 \times 10^{- 16} P a$
$E_{T} = N \frac{3}{2} k T E_{T} = 5 \times \frac{3}{2} \times 1.38 \times 10^{- 23} \times 3 P V = \frac{2}{3} N E P \times 10^{- 6} = \frac{2}{3} \times 5 \times \frac{3}{2} \times 1.38 \times 10^{- 23} \times 3 P = 2.07 \times 10^{- 16} P a$

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In a certain region of space there are only 5 molecules per cm3 on an average. The temperature is 3K. What is the average pressure of this very dilute gas?

The correct option is D 2.07×10−16PaET=N32kTET=5×32×1.38×10−23×3PV=23NEP×10−6=23×5×32×1.38×10−23×3P=2.07×10−16Pa

In a certain region of space there are only 5 molecules per cm $^{3}$ on an average. The temperature is 3K. What is the average pressure of this very dilute gas?

The correct option is D $2.07 \times 10^{- 16} P a$
$E_{T} = N \frac{3}{2} k T E_{T} = 5 \times \frac{3}{2} \times 1.38 \times 10^{- 23} \times 3 P V = \frac{2}{3} N E P \times 10^{- 6} = \frac{2}{3} \times 5 \times \frac{3}{2} \times 1.38 \times 10^{- 23} \times 3 P = 2.07 \times 10^{- 16} P a$