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iragen [17]
3 years ago
7

A balloon contains 2.3 mol of helium at 1.0 atm , initially at 240 ∘C. What's the initial volume? What's the volume after the ga

s cools at constant pressure to 14 ∘C? How much work does the gas do in this process?
Physics
1 answer:
pashok25 [27]3 years ago
5 0
A) initial volume
We can calculate the initial volume of the gas by using the ideal gas law:
p_i V_i = nRT_i
where
p_i=1.0 atm=1.01 \cdot 10^5 Pa is the initial pressure of the gas
V_i is the initial volume of the gas
n=2.3 mol is the number of moles
R=8.31 J/K mol is the gas constant
T_i=240^{\circ}C=513 K is the initial temperature of the gas

By re-arranging this equation, we can find V_i:
V_i =  \frac{nRT_i}{p_i} = \frac{(2.3 mol)(8.31 J/mol K)(513 K)}{1.01 \cdot 10^5 Pa}=0.097 m^3

2) Now the gas cools down to a temperature of
T_f = 14^{\circ}C=287 K
while the pressure is kept constant: p_f = p_i = 1.01 \cdot 10^5 Pa, so we can use again the ideal gas law to find the new volume of the gas
V_f =  \frac{nRT_f}{p_f}= \frac{(2.3 mol)(8.31 J/molK)(287 K)}{1.01 \cdot 10^5 Pa} = 0.054 m^3

3) In a process at constant pressure, the work done by the gas is equal to the product between the pressure and the difference of volume:
W=p \Delta V= p(V_f -V_i)
by using the data we found at point 1) and 2), we find
W=p(V_f -V_i)=(1.01 \cdot 10^5 Pa)(0.054 m^3-0.097 m^3)=-4343 J
where the negative sign means the work is done by the surrounding on the gas.
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A research submarine can withstand an external pressure of 62 megapascals (million pascals) all the while maintaining a comforta
Alex

Answer:

<em>The depth will be equal to</em> <em>6141.96 m</em>

<em></em>

Explanation:

pressure on the submarine P_{sea} = 62 MPa = 62 x 10^6 Pa

we also know that P_{sea} = ρgh

where

ρ is the density of sea water = 1029 kg/m^3

g is acceleration due to gravity = 9.81 m/s^2

h is the depth below the water that this pressure acts

substituting values, we have

P_{sea} = 1029 x 9.81 x h = 10094.49h

The gauge pressure within the submarine P_{g} = 101 kPa =  101000 Pa

this gauge pressure is balanced by the atmospheric pressure (proportional to 101325 Pa) that acts on the surface of the sea, so it cancels out.

Equating the pressure P_{sea}, we have

62 x 10^6 = 10094.49h

depth h = <em>6141.96 m</em>

7 0
3 years ago
At a certain time a particle had a speed of 48 m/s in the positive x direction, and 4.5 s later its speed was 92 m/s in the oppo
larisa86 [58]

Answer:

-31.1 m/s^2

Explanation:

The acceleration of an object is the rate of change of velocity of the object.

Mathematically, it is calculated as:

a=\frac{v-u}{t}

where

u is the initial velocity

v is the final velocity

t is the time taken for the velocity to change from u to v

Acceleration is a vector, so it is important to also take into account the direction of the velocity.

For the particle in this problem, we have:

u = +48 m/s is the initial velocity (positive direction)

v = -92 m/s is the final velocity (negative direction)

t = 4.5 s is the time interval

Therefore, the average acceleration is

a=\frac{v-u}{t}=\frac{-92-(+48)}{4.5}=-31.1 m/s^2

4 0
3 years ago
anonymous 4 years ago The turbidity levels of water in four locations are shown in the table below. Which of these conclusions i
Tresset [83]
Hello!

The Correct Answer would 100% be:

Option "C".

"People in location C would complain about foul taste in water".

I Hope my answer has come to your Help. Thank you for posting your question here in Brainly. We hope to answer more of your questions and inquiries soon. Have a nice day ahead! :)

(Mark As Brainliest IF Helped!)

-TheOneAboveAll :D
4 0
3 years ago
Read 2 more answers
11. Will the cube in#10 float in water? Will it float in benzene?
densk [106]
Where is the cube I don't see any picture?
5 0
3 years ago
2. A car accelerates uniformly from +10.0 m/s to +50.0 m/s over a distance of 225 m. How long did it take to go that distance? S
artcher [175]
Let's call the constant acceleration a.
At a time t, its speed will thus be v(t)=a*t+v0 where v0 is its initial speed, here 10 m/s. Hence v(t)=a*t+10.

From there we can deduce the position P(t)=a*t^2/2+10t+p0 where p0 is the initial position, here 0.

Hence P(t)=a*t^2/2+10t

Let's call T the time at which it's at 50 m/s, we know that P(T)=225m and that v(T)=50 m/s hence a*T+10=50 thus a=40/T and P(T)=(40/2+10)T=30T

Hence T=225/30=7.5

It took 7.5 seconds


7 0
2 years ago
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