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irina [24]
2 years ago
10

The gas law for a fixed mass m of an ideal gas at absolute temperature T, pressure P, and volume V is PV=mRT, where R is the gas

constant. Find the partial derivatives
Physics
1 answer:
Solnce55 [7]2 years ago
4 0

Answer:

the equation is

P*V = mRT.

The partial derivatives are:

For P = mRT/V.

dP/dV = -\frac{mRT}{V^2}

dP/dT = \frac{mR}{V}

now, if we take:

V = mRT/P

dV/dT = \frac{mR}{P}

dV/dP = -\frac{mRT}{P^2} = (dP/dV)^{-1} = -\frac{V^2}{mRT}

If we take T = PV/mR

dT/dP = \frac{V}{mR} = (dP/dT)^{-1}

dT/dV = \frac{P}{mR} = (dV/dT)^{-1}

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Record your model values in the table below
AlladinOne [14]
There not enough inform here to answer this question
8 0
3 years ago
A projectile of mass 1.800 kg approaches a stationary target body at 4.800 m/s. The projectile is deflected through an angle of
Artyom0805 [142]

Answer:

P = 5.22 Kg.m/s

Explanation:

given,

mass of the projectile = 1.8 Kg

speed of the target = 4.8 m/s

angle of deflection = 60°

Speed after collision = 2.9 m/s

magnitude of momentum after collision = ?

initial momentum of the body = m x v

                                                  = 1.8 x 4.8 = 8.64 kg.m/s

final momentum after collision

momentum along x-direction

P_x = m v cos θ

P_x = 1.8 x 2.9 x  cos 60°

P_x = 2.61 kg.m/s

momentum along y-direction

P_y = m v sin θ

P_y = 1.8 x 2.9 x  sin 60°

P_y = 4.52 kg.m/s

net momentum of the body

P = \sqrt{P_x^2 + P_y^2}

P = \sqrt{4.52^2 + 2.61^2}

 P = 5.22 Kg.m/s

momentum magnitude after collision is equal to P = 5.22 Kg.m/s

5 0
3 years ago
If the frog lands with a velocity equal to its average velocity and comes to a full stop 0.25s later, what is the frog’s average
Nina [5.8K]

Answer:

Average accelation = -4V

Explanation:

a=\frac{V-V0}{t}

V=0 m/s (because the frog stopped)

V0 = V (average velocity)

t= 0,25 s

So;

a=\frac{V-V0}{t}=\frac{0-V}{0.25}=-4V

4 0
3 years ago
Calculate the amount of heat liberated (in kj) from 411 g of mercury when it cools from 88.0°c to 12.0°c.
Katen [24]
You should note that the melting point of mercury is -38.83°C, while the boiling point is at 356.7°C. Then, that means that there is no latent heat involved here. We only compute for the sensible heat.

ΔH = mCpΔT
The Cp of mercury is 0.14 J/g·°C
Thus,
ΔH = (411 g)(0.14 J/g·°C)(88 - 12°C)
<em>ΔH = 4,373.04 J</em>
5 0
3 years ago
A cat runs at 2 m/s for 3 s and then slows to a stop with an acceleration of 0.80 m/s2. What is
sp2606 [1]

Answer:

see that the correct one is B

Explanation:

To solve this exercise let us use the kinematic relations

           v² = v₀² - 2 a x

as they indicate that the car stops, therefore the final speed is yield v = 0

          x = v₀² / 2a

let's calculate

          x = 2²/(2 0.8)

         x = 2.5 m / s²

When reviewing the answers we see that the correct one is B

7 0
2 years ago
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