Question

4. 20 moles of ideal helium gas are initially at standard temperature andpressure. The gas undergoes a change such that the pressure is doubled andthe temperature is halved. (a) What is the final volume in cubic meters? (b) lf the mass of a helium atom is 6.69 x 10^27 kg, what is the mean squared speed of the helium atoms in the final state?

Answer 1

given
no of moles, n = 20 moles
standart temperature, T1 = 0 C = 273 K
standard pressure, P1 = 1 atm = 1.0134*10^5 pa

a) given P2 = 2*P1 = 2*1.0134*10^5 pa

T2 = T1/2 = 273/2 = 136.5 K

V2 = ?

use, P2*V2 = n*R*T2

V2 = n*R*T2/P2

= 20*8.314*136.5/(2*1.0134*10^5)

= 0.112 m^3 <<<<<<-------------------Answer

b)

T2 = 136.5 K
m = 6.69*10^-27
we know, v_rms = sqrt(3*k*T/m) (here k is Boltzman's constant)

= sqrt(3*1.38*10^-23*136.5/(6.69*10^-27))

= 919 m/s <<<<<<-------------------Answer