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

Answer:

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

Explanation:

Given that,

Pressure $P=4.63\times10^{4}\ Pa$

Remove heat $\Delta U= -1.95\times10^{4}\ J$

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

$dU=-1.95\times10^{4}-(4.63\times10^{4}\times3.14\times(0.272)^2\times(-0.163))$

$dU=-17746.78\ J$

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