Question

0.014 kg of Nitrogen is enclosed in a vessel at a temperature of 27°C how much heat has to be transferred to the gas to double its rms.speed ​

Answer 1

$<b> <u> Heya!! </u> </b>$

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→ Given mass of Nitrogen = 0.014 Kg.

So let's calculate the number of moles in the given mass .

→Using the formula ,

$= > number \: of \: moles \: = \frac{given \: mass}{molecular \: mass}$

$= > n = \frac{0.014 \times 10 {}^{3} }{28} = 0.5 \: mol$

[ This is just by changing the mass into grams. And molecular mass of Nitrogen = 28 u ]

=> We know that Nitrogen is a diatomic gas ,

•°• Cv = 5/2 R

{ R = 2 for diatomic gases }

$= > cv = \frac{5}{2} \times 2 = 5$

•°• Molar heat capacity at constant volume is 5 cal.mol^-1 K^-1

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Also we know that the Root mean Square speed { rms} is directly proportional to the square root of the absolute temperatures of the gas ,

$= > \frac{v2}{v1} = \sqrt{ \frac{t2}{t1} } = 2$

•°• T2 = 4(T1)

Then , the change in temperature ∆T is given by ,

∆T = T2 - T1

=> ∆T = 4T1 - T1

•°• $<b> ∆T = 3T1 </b>$

Now , initial temperature is 27° C = 300 K ! So :-

∆T = 3 × 300

$<u><b> ∆T = 900 K </u> </b>$

Using these values , insert them in the formula for ∆q

→ ∆q = n (Cv)∆T

$= > \frac{1}{2} \times 5 \times 900$

•°• $<b> Heat required is 2250 Cals.$

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Answer 2

Mass.m.of N = 0.014 Kg.given

Atomic mass = 28

number of.mol= 0.5 mol

Cv = 5/2 R for dia atomic

Cv = 5

Use formula...

v2/v1 = t2/t1

T2 = 4(T1)

∆T = 3T1

Ti = 300 K

∆T = 3 × 300

∆T = 900 K

→ ∆q = n× (Cv)×∆T

Heat =2250 Cals.

Tq!!