About 0.014 kg nitrogen is enclosed in a vessel at temperature of `27^(@)C` How much heat has to be transferred to the gas to double the rms speed of its molecules ? `(R = 2 cal//mol K)`

About 0.014 kg nitrogen is enclosed in a vessel at temperature of `27^

1.	About 0.014 kg nitrogen is enclosed in a vessel at temperature of `27^(@)C` How much heat has to be transferred to the gas to double the rms speed of its molecules ? `(R = 2 cal//mol K)`
Answer» As gas is enclosed in the cylinder, V = constant `(Delta Q)_(v) = mu C_(v) Delta T` Here, `mu = (0.014 xx 10^(3))//28 = (1//2) mol` And as nitrogen is diatomic, `C_(v) = (5//2) R`, Futher, as according to the given problem, `((nu_(rms))_(2))/((nu_(rms))_(1)) = sqrt((T_(2))/(T_(1))) = 2`, i.e., `T_(2) = 4T_(1)` `Delta T = 4 T_(1) - T_(1) = 3T_(1) = xx 300 = 900 K` `(Delta Q)_(v) = (1)/(2) xx (5)/(2) xx 2 xx 900 = 2250 cal`