Answer: \(a + r + t = 1\); \(v_{H_2} : v_{O_2} = 8 : \sqrt{3} \approx 4.62 : 1\).
(i) Relation: Let \(Q\) be the total radiant energy (heat) incident on a body in a given time. Part \(Q_a\) is absorbed, part \(Q_r\) is reflected and part \(Q_t\) is transmitted. By conservation of energy,
\[Q = Q_a + Q_r + Q_t\]
Dividing by \(Q\):
\[1 = \frac{Q_a}{Q} + \frac{Q_r}{Q} + \frac{Q_t}{Q}\]
By definition, \(a = Q_a/Q\) is the coefficient of absorption (absorptance), \(r = Q_r/Q\) the coefficient of reflection (reflectance) and \(t = Q_t/Q\) the coefficient of transmission (transmittance). Hence
\[\boxed{a + r + t = 1}\]
For a perfectly black body \(r = t = 0\), so \(a = 1\); for an opaque body \(t = 0\), so \(a + r = 1\).
(ii) Numerical: \(v_{\text{rms}} = \sqrt{\dfrac{3RT}{M}}\), so \(v_{\text{rms}} \propto \sqrt{T/M}\).
\(T_H = 127 + 273 = 400\) K, \(M_H = 2\); \(T_O = 27 + 273 = 300\) K, \(M_O = 32\).
\[\frac{v_H}{v_O} = \sqrt{\frac{T_H}{M_H}\times\frac{M_O}{T_O}} = \sqrt{\frac{400}{2}\times\frac{32}{300}} = \sqrt{\frac{64}{3}} = \frac{8}{\sqrt{3}} \approx 4.62\]
So the r.m.s. speed of hydrogen at 127 °C is about 4.62 times that of oxygen at 27 °C, i.e. \(v_H : v_O = 8 : \sqrt{3} \approx 4.62 : 1\).
