If 15 g mass of nitrogen gas is enclosed in a vessel at a temperature 27-C- Amount of heat transfered to the gas- so that rms velocity of molecules is doubled is about-Take R-8-3 J-K mole-

Given that, nbsp; T=27+273=300 K nbsp; m=15 g M=28 g Heat transferred nbsp;in closed vessel is given by, nbsp; Q=nC_vΔ T.....(1) Rms speed is given by ...

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Standard XII

Physics

Pressure vs Velocity

If 15 g mass ...

Question

If $15 g$ mass of nitrogen gas is enclosed in a vessel at a temperature $27^{\circ} C$ . Amount of heat transfered to the gas, so that rms velocity of molecules is doubled is about
[Take $R = 8.3 J/K mole$ ]

14 kJ

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6 kJ

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0.9 kJ

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10 kJ

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Solution

The correct option is D $10 kJ$
Given that,
$T = 27 + 273 = 300 K$
$m = 15 g$
$M = 28 g$

Heat transferred in closed vessel is given by,
$Q = n C_{v} Δ T . . . . . (1)$

Rms speed is given by
$V_{r m s} = \sqrt{\frac{3 R T}{M}}$
$V_{r m s} \propto \sqrt{T}$

As Rms speed is doubled, the temperature will be quadrupled
$T^{'} = 4 T$

For diatomic gas
$γ_{d i a t o m i c} = \frac{7}{5}$
and
$C_{v} = \frac{5}{2} R$

Putting all the values in the equation $1$
$Q = n C_{v} Δ T = \frac{15}{28} \times \frac{5 R}{2} \times (4 T - T)$
$Q = \frac{15}{28} \times \frac{5 \times 8.3}{2} \times 3 \times 300$
$Q = 10000 J = 10 kJ$

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If 15 g mass of nitrogen gas is enclosed in a vessel at a temperature 27∘C. Amount of heat transfered to the gas, so that rms velocity of molecules is doubled is about [Take R=8.3 J/K mole]

If $15 g$ mass of nitrogen gas is enclosed in a vessel at a temperature $27^{\circ} C$ . Amount of heat transfered to the gas, so that rms velocity of molecules is doubled is about
[Take $R = 8.3 J/K mole$ ]