If the gas is compressed back to its original volume again in an isothermal process, what is the work done on the gas? assume there are 3.0 moles of gas at 273 k. give an answer in kj.
(a) 8.5
(b) 9.6
(c) -7.5
(d) 2.6
(d) -6.9

W= -7.5 KJ. C

Explanation:

If the gas is compressed back to its original volume again in an isothermal process, what is the work done on the gas? Assume there are 3.0 moles of gas at 273 K. Give an answer in kJ. n Ideal Gas Fills A Container. A Membrane Is Broken Which Allows The Gas To Expand Into A New Volume Which Is 3 Times As Large As The Old Volume. The Gas Is Also Cooled To Half Its Original Temperature

Explanation:

for an isothermal compression ,temperature is constant

Moreover, from the equation of state of ideal gas of isothermal process

PV = nRT1

or

p(V ) = nRT/V 2

we get

W =

Int p(V ) dV. V1 to V23

=substituting equation 2 into equation 3

W=Int nRT/V. dV

= nRT [lnV )] Within boundary =V2 and V1

= nRT lnV2/V1

Since its compressed to its original volume

v1=3V

v2=V

n=3moles

R=8.314J/molK

T=273K

3*8.314*273Lin(1/3)

W= -7.5 KJ C

The minus shows that work is done by the systemxplanation:

for an isothermal compression ,temperature is constant

Moreover, from the equation of state of ideal gas of isothermal process

PV = nRT1

or

p(V ) = nRT/V 2

we get

W =

Int p(V ) dV. V1 to V23

=substituting equation 2 into equation 3

W=Int nRT/V. dV

= nRT [lnV )] Within boundary =V2 and V1

= nRT lnV2/V1

Since its compressed to its original volume

v1=3V

v2=V

n=3moles

R=8.314J/molK

T=273K

3*8.314*273Lin(1/3)

W= -7.5 KJ C

The minus shows that work is done by the system

seeing the graph would

see the picture for the vector

the magnitude of the vector by pythagorean theorem

r = √( (-10)^2 + (3)^2 ) = √109

magnitude = √109 units

the direction angle between vector and x-axis is

tan^(-1)(3/10) = 16.699

so the direction is approximately 16.7 degrees north of west