A gas is compressed from an initial volume of 5.55 L to a final volume of 1.22 L by an external pressure of 1.00 atm. During the compression, the gas releases 124 J of heat. What is the change in the internal energy of the gas?

change in the volume of the gas = 5.55 - 1.22

= 4.33 X 10⁻³ m³

external pressure (constant) P = 1 x 10⁵ Pa

work done on the gas

=external pressure x change in volume

= 10⁵ x 4.33 X 10⁻³

=4.33 x 10²

433 J

Using the formula

Q = ΔE + W, Q is heat added, ΔE is change in internal energy, W is work done by the gas

Given

Q = - 124 J (heat is released so negative)

W = - 433 J. (work done by gas is negative, because it is done on gas)

- 124 = ΔE - 433

ΔE = 433 - 124

= 309 J

There is increase of 309 J in the internal energy of the gas.