What effects does the conductor have on the potential field?
1 answer:
You might be interested in
Answer:
1.98 atm
Explanation:
Given that:
Temperature = 28.0 °C
The conversion of T( °C) to T(K) is shown below:
T(K) = T( °C) + 273.15
So,
T₁ = (28 + 273.15) K = 301.15 K
n = 1
V = 0.500 L
Using ideal gas equation as:
PV=nRT
where,
P is the pressure
V is the volume
n is the number of moles
T is the temperature
R is Gas constant having value = 0.0821 L atm/ K mol
Applying the equation as:
P × 0.500 L = 1 ×0.0821 L atm/ K mol × 301.15 K
⇒P (ideal) = 49.45 atm
Using Van der Waal's equation
R = 0.0821 L atm/ K mol
Where, a and b are constants.
For Ar, given that:
So, a = 1.345 atm L² / mol²
b = 0.03219 L / mol
So,
⇒P (real) = 47.47 atm
Difference in pressure = 49.45 atm - 47.47 atm = 1.98 atm
Answer:
80⁰
Explanation:
triangle AOB = 180
So, a = 50
triangle CAB = a+a+x=180
CAB = 50+50+X=180
CAB = X = 180-100
X=80
HOPE IT HELPS
