Answer:
Explanation:
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.
