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Margarita [4]
3 years ago
6

Two moles of an ideal monatomic gas are at a temperature of 345 K. Then, 2250 J of heat is added to the gas, and 870 J of work i

s done on it. What is the final temperature of the gas
Physics
1 answer:
Lapatulllka [165]3 years ago
5 0

Answer: The final temperature is 470K

Explanation: Using the relation;

Q= ΔU +W

Given, n = 2mol

Initial temperature T1= 345K

Heat =Q= 2250J

Workdone=W=-870J(work is done on gas)

T2 =Final temperature =?

ΔU =3/2nR(T2-T1)

ΔU=3/2 × 2 ×8.314 (T2 - 345)

ΔU=24.942(T2-345)

Therefore Q = 24.942(T2-345)+ (-870)

2250=24.942(T2-345)+ (-870)

125.09=(T2-345)

T2 =470K

Therfore the final temperature is 470K

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Officials begin to release water from a full man-made lake at a rate that would empty the lake in 4 weeks, but a river that can
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Answer:

5 weeks and 5 days is required to empty the lake

Explanation:

Officials begin the remove water from a full man made lake

The lake can be emptied in 4 weeks

= -1/4

A river can fill the lake up in 15 weeks

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Let t represent the number of weeks that is required to empty the lake

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Therefore the number of weeks it takes to empty the lake can be calculated as follows

-1/t= -1/4 + 1/15

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-11×t= -1×60

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6 0
3 years ago
Suppose the battery in a clock wears out after moving thousand coulombs of charge through the clock at a rate of 0.5 Ma how long
Ksivusya [100]

Answer:

Hello your question is poorly written below is the complete question

Suppose the battery in a clock wears out after moving Ten thousand coulombs of charge through the clock at a rate of 0.5 Ma how long did the clock run on does battery and how many electrons per second slowed?

answer :

a) 231.48 days

b) n = 3.125 * 10^15

Explanation:

Battery moved 10,000 coulombs

current rate = 0.5 mA

<u>A) Determine how long the clock run on the battery. use the relation below</u>

q = i * t ----- ( 1 )

q = charge , i = current , t = time

10000 = 0.5 * 10^-3 * t

hence  t = 2 * 10^7 secs

hence the time = 231.48  days

<u>B) Determine how many electrons per second flowed </u>

q = n*e ------ ( 2 )

n = number of electrons

e = 1.6 * 10^-19

q = 0.5 * 10^-3 coulomb ( charge flowing per electron )

back to equation 2

n ( number of electrons ) = q / e = ( 0.5 * 10^-3 ) / ( 1.6 * 10^-19 )

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