All categories

3 years ago

A 2 L balloon filled with gas is warmed from 280 K to 700 K. What is the volume of the gas after it is heated?

Physics

1 answer:

irakobra [83]3 years ago

7 0

Answer:

New volume, v2 = 0.8L

Explanation:

<u>Given the following data;</u>

Original Volume = 2L

Original Temperature = 280K

New Temperature = 700K

To find new volume V2, we would use Charles' law.

Charles states that when the pressure of an ideal gas is kept constant, the volume of the gas is directly proportional to the absolute temperature of the gas.

Mathematically, Charles is given by;

$VT = K$

$\frac{V1}{T1} = \frac{V2}{T2}$

Making V2 as the subject formula, we have;

$V_{2}= \frac{V1}{T1} * T_{2}$

$V_{2}= \frac{2}{700} * 280$

$V_{2}= 0.0029 * 280$

V2 = 0.8L

Therefore, the volume of the gas after it is heated is 0.8L.

You might be interested in

Determine the voltage ratings of the high-and-low voltage windings for this connection and the MVA rating of the autotransformer

SIZIF [17.4K]

Complete Question

The complete question is shown on the first uploaded image

Answer:

The High Voltage Rating for Auto - Transformer is 86kV

The Low Voltage Rating for Auto - Transformer is 78kV

The MVA rating is 268.75 $MVA$

The efficiency is 99.4%

Explanation:

From the question we are given are given that

The transformer has Mega Volt Amp rating of 25MVA

The frequency is 60-Hz

Voltage rating 8.0kV : 78kV

The short circuit test gives : 453kV,321A,77.5kW

The open circuit test gives : 8.0kV, 39.6A, 86.2kW

This can be represented on a diagram shown on the second uploaded image

From this diagram we can deduce that the The High Voltage Rating for Auto - Transformer is 86kV and the Low Voltage Rating for Auto - Transformer is 78kV

Now to obtain the current flowing through the 8kV coil in the Auto-transformer we have

$\frac{25 \ Mega \ Volt\ Ampere }{8\ Kilo Volt}$

The volt will cancel each other

$\frac{25*10^6}{8*10^3} = 3125\ A$

Now to obtain the required MVA rating we would multiply the value of Power obtained during the open circuit test by the value of the current calculated.we are making use of the power obtain during open circuit testing because the transformer at this point is not under any load.

$MVA \ rating = (86*10^3)(3125) =268.75$

We need to understand that Iron losses is due to open circuit test which has power = 86.2kW

While copper loss is due to short circuit test which has power = 77.5kW

The the current flowing through the secondary coil $I_2$ as shown in the circuit diagram can be obtained as

$I_2 = \frac{25*10^6}{78*10^3} =320.52 A \approx 321$

Now the efficiency can be obtained as thus

$\frac{(operational \ MVA )*(Power factor \pf))}{(operational\ MVA (power factor pf) + copper loss + Iron loss)}*\frac{100}{1}$

=99.941%

8 0

2 years ago

My school doesn’t really teach help

Olegator [25]

C
Unbalanced causes something to move because the net force is greater than zero

7 0

3 years ago

In a candy factory, the nutty chocolate bars contain 20.0 % pecans by mass. If 4.0 kg of pecans were used for candy last Tuesday

ladessa [460]

Answer:

44.09 pounds

Explanation:

We got that 20 % of the mass of a nutty chocolate bar its pecans, if 4.0 kg of pecans were used, we need to find the X in the equation

$0.2 * X = 4.0 \ kg$

where X its the total mass of nutty chocolate bars produced. So, we can just divide by 0.2 on both sides, and we find:

$X = 4.0 kg / 0.2$

$X = 20.0 \ kg$

Of course, we need the total mass produced in pounds, and not in kilograms. Looking at an conversion table, we can find that 1 kg its 2.20462 pounds, multiplying the value for total mass produced by the conversion factor we get:

$X = 20.0 \ kg * 2.20462 \ \frac{pounds}{kg}$

$X = 44.0924 \ pounds$

Now, we just need to round off to two significant figures. This is:

$X = 44.09 \ pounds$ ,

the total mass of nutty chocolate bars made last Tuesday to two significant figures.

5 0

3 years ago

A gang of robbers is escaping across city roofs at night. They come to the edge of one building and need to drop down to their g

REY [17]

Answer:

a) They will hit the ground with a speed of 19.6 m/s.

b) They are at a height of 20 m.

c) It is not a safe jump.

Explanation:

Hi there!

a) The equations of height and velocity in function of time of a free falling body are the following:

h = h0 + v0 · t + 1/2 · g · t²

v = v0 + g · t

Where:

h = height of the object at time t.

h0 = initial height.

v0 = initial velocity.

t = time.

g = acceleration due to gravity (-9.8 m/s² considering downward as negative direction).

v = velocity of the object at time t.

Using the equation of velocity, let's find the velocity at which they will hit the ground. The pebble is dropped (initial velocity = 0) and it takes 2 s to reach the ground:

v = v0 + g · t (v0 = 0)

v = g · t

v = -9.8 m/s² · 2.0 s

v = -19.6 m/s

They will hit the ground with a speed of 19.6 m/s.

b)Now, we have to use the equation of height:

h = h0 + v0 · t + 1/2 · g · t²

If we place the origin of the frame of reference on the ground, we have to find the initial height (h0) knowing that at t = 2.0 s, h = 0 m

0 m = h0 - 1/2 · 9.8 m/s² · (2.0 s)²

h0 = 1/2 · 9.8 m/s² · (2.0 s)²

h0 = 20 m

They are at a height of 20 m.

c)According to a NASA paper (Issues on Human Acceleration Tolerance After Long-Duration Space Flights, figure 10), if you fall with a vertical velocity greater than 17 m/s it is unlikely that you will survive. So, it is not a safe jump.

3 0

2 years ago

Find the acceleration a body whose velocity increases from 11m/s to 33m/s in 10 seconds

solong [7]

Answer:

1.1 m/(s)^2

Explanation:

u=11 m/s

v=33 m/s

t=10s

v=u+at

=> 33=22+(a)(10)

=> 33-22=10a

=> 10a=11

=> a=11/10=1.1 m/(s)^2

7 0

2 years ago