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Physics - Heat & Thermo Question with Solution | TestHub

PhysicsHeat & ThermoKTGMedium2 minPYQ_2019

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A $15 g$ mass of nitrogen gas is enclosed in a vessel at a temperature, $27^{o} C$ . The amount of heat transferred to the gas, so that $R . M . S .$ velocity of molecules is doubled, is about.
$[R = 8 .3 J {(K mole)}^{- 1}]$

Options:

Answer:

Solution:

The formula of root-mean-square speed, $v_{r m s} = \sqrt{\frac{3 R T}{M}}$ . For doubling the $v_{r m s},$ temperature, should be made $4$ -time is, $T_{f} = (273 + 27) \times 4 K = 1200 K$ , the formula of, $∆ Q = ∆ U + W (W = 0)$ ,
$\Rightarrow ∆ Q = ∆ U$ $= n C_{V} ∆ T$ $= \frac{15}{28} \times \frac{5}{2} R ∆ T$
$\Rightarrow ∆ Q = \frac{15}{28} \times \frac{5}{2} \times 8.3 \times (1200 - 300) J$ $\approx 10 KJ .$

Stream:JEESubject:PhysicsTopic:Heat & ThermoSubtopic:KTG

⏱ 2mℹ️ Source: PYQ_2019

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