Correct answer is (12)

Because the vessel is closed, it will be an isochoric process.

To double the speed, temperature must be 4 times (v $\alpha$$\sqrt{T}$)

So, T_f = 1600 K, T_i = 400 K

number of moles are $\frac{56}{28}=2$

so Q = nCv $\mathrm{\Delta }$T = 2 $\times$ $\frac{5}{2}$ $\times$ 2 $\times$ 1200

= 12000 cal = 12 K cal

Correct answer is (A)

$V_{rms}=\sqrt{\frac{3KT}{M}}$

$\frac{{(V_{rms})}_{O_2}}{{(V_{rms})}_{H_2}}=\sqrt{\frac{M_{H_2}}{M_{O_2}}}=\sqrt{\frac{2}{32}}$

$(V_{rms}{)}_{H_2}=4\times (V_{rms}{)}_{O_2}$

$=4\times 160$

$=640$ m/s

Correct answer is (A)

$V_{RMS}=\sqrt{\frac{3RT}{M_W}}$

At the same temperature $V_{RMS}\propto \frac{1}{\sqrt{M_W}}$

$\Rightarrow$ V_H > V_O > V_C

Option (a)

Correct answer is (C)

rms velocity of gas molecules is given as

$v_{rms}=\sqrt{\frac{3RT}{m}}$ ..... (i)

where, m = molar mass of the gas in kilograms per mole,

R = molar gas constant,

and T = temperature in kelvin.

According to question,

m_A < m_B < m_C

From Eq. (i),

$v_{rms}\propto \frac{1}{\sqrt{m}}$

$\therefore$ We can write,

v_A > v_B > v_C or $\frac{1}{v_A}<\frac{1}{v_B}<\frac{1}{v_C}$

Correct answer is (A)

$V_{rms}=\sqrt{\frac{3RT}{M}}$

$V_{avg}=\sqrt{\frac{8}{\pi }\frac{RT}{M}}$

$\frac{V_{rms}}{V_{avg}}=\sqrt{\frac{3\pi }{8}}$