Answer:
the work is done by the gas on the environment -is W= - 3534.94 J (since the initial pressure is lower than the atmospheric pressure , it needs external work to expand)
Explanation:
assuming ideal gas behaviour of the gas , the equation for ideal gas is
P*V=n*R*T
where
P = absolute pressure
V= volume
T= absolute temperature
n= number of moles of gas
R= ideal gas constant = 8.314 J/mol K
P=n*R*T/V
the work that is done by the gas is calculated through
W=∫pdV= ∫ (n*R*T/V) dV
for an isothermal process T=constant and since the piston is closed vessel also n=constant during the process then denoting 1 and 2 for initial and final state respectively:
W=∫pdV= ∫ (n*R*T/V) dV = n*R*T ∫(1/V) dV = n*R*T * ln (V₂/V₁)
since
P₁=n*R*T/V₁
P₂=n*R*T/V₂
dividing both equations
V₂/V₁ = P₁/P₂
W= n*R*T * ln (V₂/V₁) = n*R*T * ln (P₁/P₂ )
replacing values
P₁=n*R*T/V₁ = 2 moles* 8.314 J/mol K* 300K / 0.1 m3= 49884 Pa
since P₂ = 1 atm = 101325 Pa
W= n*R*T * ln (P₁/P₂ ) = 2 mol * 8.314 J/mol K * 300K * (49884 Pa/101325 Pa) = -3534.94 J
Answer:
T = 712.9 N
Explanation:
First, we will find the speed of the wave:
v = fλ
where,
v = speed of the wave = ?
f = frequency = 890 Hz
λ = wavelength = 0.1 m
Therefore,
v = (890 Hz)(0.1 m)
v = 89 m/s
Now, we will find the linear mass density of the wire:
where,
μ = linear mass density of wie = ?
m = mass of wire = 90 g = 0.09 kg
L = length of wire = 1 m
Therefore,
μ = 0.09 kg/m
Now, the tension in wire (T) will be:
T = μv² = (0.09 kg/m)(89 m/s)²
<u>T = 712.9 N</u>
Answer:
A downward sloped line means the object is returning to the starting point.
