**Answer:**

the local atmospheric pressure is **93.63 kPa**

the mass of the weights is **156.9 kg**

**Explanation:**

Given that;

Initial pressure of gas = 100 kPa

mass of piston = 10 kg and diameter = 14 cm = 0.14 m

g = 9.81 m/s²

Now,

P_gas = P_atm + P_piston

100 = P_atm + P_piston --------- let this equation 1

P_piston = M_piston × g / A = (10 × 9.81) / π/4×d²

P_piston = 98.1 / (π/4×( 0.14 )²)

P_piston = 98.1 / 0.01539 = 6374,269 Pa = 6.37 kPa

now, from equation 1

100 = P_atm + P_piston

we substitute

100 = P_atm + 6.37

P_atm = 100 - 6.37

P_atm = **93.63 kPa**

Therefore, the local atmospheric pressure is **93.63 kPa**

Now for pressure of the gas in the cylinder ⇒ 2×initial pressure

Pgas_2 = 2 × 100 = 200 kPa

Pgas_2 = P_atm + P_piston + P_weight

Pgas_2 = P_gas + P_weight

we substitute

200 kPa = 100 kPa + P_weight

P_weight = 200 kPa - 100 kPa

P_weight = 100 kPa = 100,000 Pa

Also;

P_weight = M×g / A

100,000 Pa = ( M × 9.81 ) / (π/4 × (0.14)²)

100,000 × 0.01539 = M × 9.81

1539 = M × 9.81

M = 1539 / 9.81

M = **156.9 kg**

Therefore, the mass of the weights is **156.9 kg**