Question #9a548

1 Answer
Feb 8, 2018

The change in internal energy is of 1061 J.

Explanation:

The internal energy variation #DeltaU# is related to the system's heat #Q# by:

#DeltaU = Q - W#, where #W# is the work done by/on the system.

Since the pressure does not vary, #W# can be simply calculated as

#W = pDeltaV#; where #p# and #DeltaV# stand for the pressure and the system's volume variation, respectively.

Putting the numerical values into our expression for #W# gives

#W = 1.5 * 10^5 Pa * (2.2 - 7.5) * 10^-3 m^3#;

#W = -7.95 * 10^2 J#.

Since we have now both #Q# and #W#, we can calculate #DeltaU#:

#DeltaU = Q - W#;

#DeltaU = 266 J - (-795)J#;

#DeltaU = 1061 J#.