STANDARD Physics First law of thermodynamics MCQ Question
A gas is allowed to expand in a well insulated container against a constant external pressure of 2.5 atm from an initial volume of 2.50 L to a final volume of 4.50 L. The change in internal energy ΔU of the gas in joules will be
−500 J
−500 J
+500 J
1136.25 J
Correct Answer
Detailed Explanation
The work done is w = −PₑₓₜΔV = −2.5(4.50 − 2.50) = −5 L atm = −5 × 101.325 J = −506.625 J. Since the container is insulated, q = 0, hence, ΔU = w = −506.625 J.
Found an issue with this question?
Related Questions
More from
What is the enthalpy change associated with the atomization of dihydrogen (H2)?
For a strongly endothermic reaction, what is the expected value of the equilibrium constant K?
For an adiabatic expansion of an ideal gas, if the initial internal energy is 500 J and the work done on the system is -200 J, what is the final inter...