Amanometer consists of a u-shaped tube containing a liquid. one side is connected to the apparatus and the other is open to the atmosphere. the pressure p inside the apparatus is given p = pex + rhogh, where pex pressure, rho is the mass density of the liquid in the tube, g = 9.806 m s−2 is the external is the acceleration of free fall, and h is the difference in heights of the liquid in the two sides of the tube. (the quantity rhogh is the hydrostatic pressure exerted by a column of liquid.) (i) suppose the liquid in a manometer is mercury, the external pressure is 760 torr, and the open side is 10.0 cm higher than the side connected to the apparatus. what is the pressure in the apparatus? the mass density of mercury at 25 °c is 13.55 g cm−3 . (ii) in an attempt to determine an accurate value of the gas constant, r, a student heated a container of volume 20.000 dm3 filled with 0.251 32 g of helium gas to 500 °c and measured the pressure as 206.402 cm in a manometer filled with water at 25 °c. calculate the value of r from these data. the mass density of water at 25 °c is 0.997 07 g cm−3

1) The pressure in the apparatus is 101311.7 Pascal = 759.9 torr

2) The value of R is 0.0824 L*atm/mole*K

Explanation:

1) In a U-tube manometer, the pressure difference across both ends is equal to the pressure due to height difference in the liquid at both ends with more pressure at the end where the liquid has risen.

p2-p1 =  ρ*g*h

with P1 = pressure of the liquid on the half where  the liquid has risen

with P2 = pressure of the liquid on the other half

with ρ = the density of the liquid

with g = the accelaration due to gravity

with h = the height difference at both ends

Because 1 side is open to the atmosphere

P1 = atmospheric pressure = 760 torr = 101325 Pascal

ρ = 13.55 g/cm³

g = 9.806 m/s²

h = -0.10 meter

P2 =  P1 + ρ*g*h

P2 = 101325 + 13.55*9.806 * -0.10

P2 = 101311.7 Pascal = 759.9 torr

The pressure in the apparatus is 101311.7 Pascal = 759.9 torr

b)

The ideal gas law says: R = PV/nT

with P = 206.402 cm H2O

with V = 20.00 dm³ x (1 L / 1 dm³) = 20.00 L

with n = 0.25132g He x (1 mole / 4.0026 g ) = 0.062789 mole He

with T = 500 C = 273 + 500 = 773 Kelvin

We know the following formula: P = m*g*h

If we compare pressure in mmHg to cm H2O

(ρ*g*h)Mercury = (ρ*g*h)water

⇒ g Mercury = g Water

This gives us:

(ρ*h)Mercury = (ρ*h)water

height Mercury = (ρ*h)water  / ρMercury

Height Mercury = 0.997g/cm³ * 206.402 cm H2O  / (13.55 g/cm³) x (10 mm / 1 cm) = 15.19 cm = 151.9 mmHg

Since 760 mmHg = 1 atm

151.9 /760.000 mmHg = 0.1999 atm

If we insert all in the ideal gas law P*V=nRT

R = P*V/n*T

R = (0.1999 atm) x (20.000 L) / ((0.062789 mole He) x (773K))

R = 0.0824 L*atm/mole*K

The value of R is 0.0824 L*atm/mole*K

when the volume of gas is   changed from   717.73 ml   to 852 ml   the temperature will   change   from 315   c to 452   c

    calculation

this is calculated using the charles law formula

that is,   v1/t1 = v2/t2

          v1=?

        t1 = 315 c   into   kelvin = 315 +273 = 588 k

        v2 = 852   ml

        t2 = 425c + 273 =698 k

make v1   the subject of the formula ( by   multiplying both side by t1)

  v1=( v2 xt1) / t2

v1= (852 ml   x 588 k)/ 698 k= 717.73 ml