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antiseptic1488 [7]
3 years ago
9

During the expansion of a gas in piston-cylinder the relation between the pressure and the volume is given as

Physics
1 answer:
Iteru [2.4K]3 years ago
8 0

Answer: The relationship between the pressure and the volume is given as P=K/V or PV= K. (Pressure increases as Volume decreases).

Explanation:

When gas is trapped in an enclosed medium for example a cylinder by a piston. When the piston is pushed in, the gas particles will have less room to move as the VOLUME the gas occupies has been DECREASED. This will automatically lead to frequent collision of the has particles with the wall of the container. The collision exerts forces which leads to INCREASED PRESSURE.

This obeys the Boyle's law which states that the volume occupied by a fixed mass of gas is inversely proportional to the pressure, provided temperature is kept constant. This can be mathematically expressed below:

PV=K, where K is constant

P1 V1=P2 V2

Where,

P1= initial pressure

V1= initial Volume

P2= final pressure

V2= final Volume.

Therefore, increasing pressure from initial pressure to final pressure means that initial volume will change to final volume, providing the temperature remains constant.

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A box of volume V has a movable partition separating it into two compartments. The left compartment contains 3000 particles, the
storchak [24]

Answer:

a) V1 = 4V - V2/3 and V2 = 4V -  3V1

b) Δe = 4000V - 4000V2 + 9000V1

Explanation:

Let V represent volume of the box containing the two compartments

V1 represents compartment of the left compartment

V2 represents compartment of the right compartment

Momentum of the compartments before impact:

3000V1 + 1000V2

Momentum of the compartments after impact:

V(3000 + 1000) = 4000V

a) To obtain the volume of each compartment, that is, V1 and V2, we say:

Momentum before impact = Momentum after impact

3000V1 + 1000V2 = 4000V

∴ V1 = 4000V - 1000V2/3000 = 4V - V2/3

Also, V2 = 4000V - 3000V1/1000 = 4V - 3V1

b) Change in entropy,Δe = 4000V1 - 1000V2

By substituting the V1 and V2, we have:

4000(4V - V2)/3 - 1000(4V - 3V1)

16000V - 4000V2/3 - 4000V + 3000V1

16000V -  4000V2 - 12000V + 9000V1

∴ Δe = 4000V - 4000V2 + 9000V1

6 0
3 years ago
A single-phase electrical load draws 600KVA at 0.6 power factor lagging. a) Find the real and reactive power absorbed by the loa
Whitepunk [10]

Answer:

(a). The reactive power is 799.99 KVAR.

(c). The reactive power of a capacitor to be connected across the load to raise the power factor to 0.95 is 790.05 KVAR.

Explanation:

Given that,

Power factor = 0.6

Power = 600 kVA

(a). We need to calculate the reactive power

Using formula of reactive power

Q=P\tan\phi...(I)

We need to calculate the \phi

Using formula of \phi

\phi=\cos^{-1}(Power\ factor)

Put the value into the formula

\phi=\cos^{-1}(0.6)

\phi=53.13^{\circ}

Put the value of Φ in equation (I)

Q=600\tan(53.13)

Q=799.99\ kVAR

(b). We draw the power triangle

(c). We need to calculate the reactive power of a capacitor to be connected across the load to raise the power factor to 0.95

Using formula of reactive power

Q'=600\tan(0.95)

Q'=9.94\ KVAR

We need to calculate the difference between Q and Q'

Q''=Q-Q'

Put the value into the formula

Q''=799.99-9.94

Q''=790.05\ KVAR

Hence, (a). The reactive power is 799.99 KVAR.

(c). The reactive power of a capacitor to be connected across the load to raise the power factor to 0.95 is 790.05 KVAR.

8 0
3 years ago
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