Dear Student, Side of cubical vessel, l = 5 cmVelocity in x-direction, Vx = 105 cm/secSo, Change in momentum = mVx-(-mVx) = 2mVxwhere, m is the mass of moleculeNumber if collisions of molecule per second = Vx2lSo, rate of change in momentum in x-direction = (2mVx)(Vx2l) =mVx2lmass of He molecule = 4 g (as it is monoatomic gas)Therefore, rate of change in momentum in x-direction =4×(105)25 = 4×10105=8×109 gcm/sec

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Dear Student,

Side of cubical vessel, l = 5 cm Velocity in x - direction, V_{x} = 10^{5} cm / \sec So, Change in momentum = {mV}_{x} - (- {mV}_{x}) = 2 {mV}_{x} where, m is the mass of molecule Number if collisions of molecule per second = \frac{V_{x}}{2 l} So, rate of change in momentum in x - direction = (2 {mV}_{x}) (\frac{V_{x}}{2 l}) = \frac{{mV}_{x}^{2}}{l} mass of He molecule = 4 g (as it is monoatomic gas) Therefore, rate of change in momentum in x - direction = \frac{4 \times (10^{5})^{2}}{5} = \frac{4 \times 10^{10}}{5} = 8 \times 10^{9} gcm / \sec

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