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 ????


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Answer 1

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.


$Tq$

Answer 2

Answer:

The heat has to be transferred to the gas, is equal to 2250Calries.

Explanation:

To find heat transferred:    Δ$Q_{v}=nC_{v}$ΔT               ...........(1)

Consider that n is the number of moles of nitrogen gas.

The weight of N₂ gas  $=0.014*10^{3}g$

Then  $n=\frac{0.014*10^{3}}{28} =0.5 mol$

The Heat capacity of N₂ gas, $C_{v}=\frac{5}{2} R$     where R= 2calK⁻¹mol⁻¹

As we know that $V_{r.m.s.}$ are directly proportional to $\sqrt{T}$

$\frac{(V_{r.m.s.})_{2}}{V_{r.m.s.})_{1}} =\sqrt{\frac{T_{2}}{T_{1}} } =2$  

Squaring both sides, we get

$T_{2}=4T_{1}$

Δ$T= T_{2}-T_{1}= 4T_{1}-T_{1} = 3T_{1}=(3)(300) =900K$

Put all the values in eq.(1),

Δ$Q_{v}=(0.5mol)(\frac{5}{2} )(2 calK^{-1}mol^{-1})(900K)=2250cal$

Therefore the heat transferred to nitrogen gas to double root mean square speed is 2250cal.