How Much Heat Is Needed For Process 1 2

Physics: Questions about monoatomic gases? ۔

Hello friends

I have a problem I can't solve. Related image is here:

At point 1:

P = 3 atm = 3 x 1.01 x 10 Pa = 3.03 x 10 Pa.

V = 100cm³ = 100x10Â »Ã Â m³ = 1x10Â» Â´ m³

T = 100 ° C = 100 + 273 K = 373 K.

Therefore, we can find the amount of gas:

PV = nRT

3.03x10Â x 1x10Â »ÃÂ´ = n x 8.31 x 373.

n = 9,775x10 Â ol ol ol mole.

________________________________________________

At point 2, for example, we can determine the new temperature.

PÃ Â, GO, Â / TÃ Â, Â = PÃ Â, GO, Â / TÃ Â, Â,

Since PÃ Â, Â = PÃ Â, GO, Â / TÃ Â, Â = GO, Â, / TÃ Â,

TÃ Â, Â = (GO, Â, / GO, Â) T Â,.

. . = (300/100) x 373.

. . = 1119K

________________________________________________

Calculate the temperature from 1 to 2:

The specific temperature at constant pressure CÂº = (5/2) R, ie the heat supplied is:

Q = nCÃ T ”T

. . = 9.775x10 Â »ÃÂ³ x (5/2) x8.31 x (1119 373)

. . = 151.5J

____________________________

Calculate the temperature from 2 to 3:

It happens at a constant volume. The specific temperature at constant temperature CáÂµÂ = (3/2) R is therefore the given temperature:

Q = nCáÂµÂ ”T

. . = 9.775x10 Â »ÃÂ³ x (3/2) x8.31 x (373 1119)

. . = 90.9 years

(This is negative because a negative change in temperature causes heat to be lost.)

(But check out my math.)

How Much Heat Is Needed For Process 1 2

Physics: Questions about Atomic Gases? 3
Hello friends
I have a problem I can't solve. Related image is here:
Up to point 1:
P = 3 atm = 3 x 1.01 x 10 Pa = 3.03 x 10 Pa
V = 100cm³ = 100x10Â »Ã Â¶ m³ = 1x10Â» Â´ m³
T = 100 ° C = 100 + 273 K = 373 K.
So we can find the amount of gas:
PV = nRT
3.03x10 Âµ x 1x10Â »Ã Â´ = n x 8.31 x 373
n = 9,775x10 Â³ mol
________________________________________________
At point 2 we can define a new temperature, for example
PÃ, GO, Â / TÃ, Â = PÃ Â, GO, Â / TÃ Â, Â,
Since PÃ, Â = PÂ, GÃ, Â / TÃ, Â = GÃ Â, Â, / TÃ Â,
TÃ Â, Â = (GO, Â, / GO, Â) TÂ,
. . = (300/100) x 373
. . = 1119K
________________________________________________
Calculate the temperature required to go from 1 to 2:
The specific heat at constant pressure CÃ Â ° = (5/2) R, so the heat supplied is:
Q = nCÃ‚ÂšÃŽÂ ”T
. . = 9,775x10 Â »ÃÂ³ x (5/2) x8.31 x (1119 373)
. . = 151.5J
_______________
Calculate the temperature required to go from 2 to 3:
It happens at a constant volume. The specific heat at constant volume Cáµ = (3/2) R, therefore, the heat supplied is:
Q = nCáÂµÂ ”T
. . = 9,775x10 Â »ÃÂ³ x (3/2) x8.31 x (373 1119)
. . = 90.9 years
(This is negative because a negative temperature change causes the heat to dissipate.)
(But check my math.)