RMS velocity of one mole of an ideal gas was measured at different temperatures. A graph of ( u _ rms )^2 (on y -axis) and T ( K ) (on x -axis) gave straight line passing through the origin and its slope is 249 ~m ^2 ~s ⁻² ~K ⁻¹ . What is the molar mass (in . kg mol ⁻¹ ) of ideal gas? ( R =8.3 ~J ~mol ⁻¹ ~K ⁻¹ )

Step-by-step solution

RMS velocity (v_ rms )= 3 R T M aligned & v_ rms ^2= 3 RT M & y = mx + C aligned where, aligned & y = v _ rms ^2 & x = T aligned aligned & m ( slope )= 3 R M & C( intercept )=0 aligned Slope ( m )= 3 R M [ R =83 ~J ~mol ⁻¹ ~K ⁻¹ ] 3 R M =249 ~m ^2 ~s ⁻² ~K ⁻¹ M = 3 8.3 ~J ~mol ⁻¹ ~K ⁻¹ 249 ~m ^2 ~s ⁻² ~K ⁻¹ [1 ~J =1 ~kg ] M = 3 8.3 ~kg ~m ^2 ~s ⁻² ~mol ⁻¹ ~K ⁻¹ 249 ~m ^2 ~s ^2 ~K ⁻¹ [1 ~J =1 ~kg ~m ^2 ~s ] aligned & M = 24.9 ~kg ~mol ⁻¹ 249 & M =0.1 ~kg ~mol ⁻¹ & M (molar mass) =1 10⁻¹ ~kg ~mol ⁻¹ aligned