Question

A cylinder which is in a horizontal position contains an unknown noble gas at 4.63 × 104 Pa and is sealed with a massless piston. The piston is slowly, isobarically moved inward 0.163 m, while 1.95 × 104 J of heat is removed from the gas. If the piston has a radius of 0.272 m, calculate the change in internal energy of the system.

Answer 1

The change in internal energy of the system is -17746.78 J

Step-by-step explanation:

Given that,

Pressure
`P=4.63*10^(4)\ Pa`

Remove heat
`\Delta U= -1.95*10^(4)\ J`

Radius = 0.272 m

Distance d = 0.163 m

We need to calculate the internal energy

Using thermodynamics first equation

`dU=Q-W`...(I)

Where, dU = internal energy

Q = heat

W = work done

Put the value of W in equation (I)

`dU=Q-PdV`

Where, W = PdV

Put the value in the equation

`dU=-1.95*10^(4)-(4.63*10^(4)*3.14*(0.272)^2*(-0.163))`

`dU=-17746.78\ J`

Hence, The change in internal energy of the system is -17746.78 J