How to correct dimensions of capillary rise of liquid in a tube
1 answer:
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π=iMRT
Where, π is Osmotic pressure,
i=1 for non-electrolytes,
M is molar concentration of dissolved species (units of mol/L)
R is the ideal gas constant = 0.08206 L atm mol⁻¹K⁻¹,
T is the temperature in Kelvin(K),
Here, to calculate M convert into standard units mg tog, ml to L, c to Kelvin
M= ( *10⁻³ )/ 0.175 =(5.987 *10⁻⁵)mol / 0.175L = 34.21*10⁻⁵ mol/L
π=iMRT=(1)*(34.21*10⁻⁵)*(0.08206)*(298.15)=837×10⁻⁵= 8.37×10⁻³ atm
=6.36 torr
(1 atm=760 torr, 1 Kelvin =273.15 °C, 1L=1000ml, 1g=1000mg)
Answer:
Explanation:
Total volume of the shell on which charge resides
= 4/3 π ( R₁³ - R₂³ )
= 4/3 X 3.14 ( 33³ - 20³) X 10⁻⁶ m³
= 117 x 10⁻³ m³
Charge inside the shell
-117 x 10⁻³ x 1.3 x 10⁻⁶
= -152.1 x 10⁻⁹ C
Charge at the center
= - 60 x 10⁻⁹ C
Total charge inside the shell
= - (152 .1 + 60 ) x 10⁻⁹ C
212.1 X 10⁻⁹C
Force between - ve charge and proton
F = k qQ / R²
k = 9 x 10⁹ .
q = 1.6 x 10⁻¹⁹ ( charge on proton )
Q = 212.1 X 10⁻⁹ ( charge on shell )
R = 33 X 10⁻² m ( outer radius )
F =
F = 2.8 X 10⁻¹⁵ N
This force provides centripetal force for rotating proton
mv² / R = 2.8 X 10⁻¹⁵
V² = R X 2.8 X 10⁻¹⁵ / m
= 33 x 10⁻² x 2.8 x 10⁻¹⁵ /( 1.67 x 10⁻²⁷ )
[ mass of proton = 1.67 x 10⁻²⁷ kg)
= 55.33 x 10¹⁰
V = 7.44 X 10⁵ m/s