Small issue regarding the wording of a thermodynamics question

[dregozo]

Homework Statement

The issue is the word 'increase' in c)iii), where I'm pretty sure the wording should have been 'decrease' or at least 'change', since both the work done and the heat removed have a negative value.

Homework Equations

dU = dQ + dW (1st law)

The Attempt at a Solution

My answer to c)iii) was -160J.

 

[Bystander]

The issue is the word 'increase' in c)iii), where I'm pretty sure the wording should have been 'decrease' or at least 'change', since both the work done and the heat removed have a negative value.

"Increase" IS change; don't get your knickers in a knot.

 

[dregozo]

Can't agree with you there, an answer of -160J can't possibly suit the description of an INCREASE in internal energy!

 

[Chestermiller]

Something seems very wrong with this problem statement. If the initial and final states of the system are thermodynamic equilibrium states for an ideal gas, then we must have that $$\Delta (PV)=nR\Delta T$$ and
$$\Delta U=nC_v\Delta T$$Combining these two equations gives:$$\Delta U=\frac{C_v}{R}\Delta (PV)$$For a monoatomic ideal gas the ratio of the heat capacity to the gas constant is 1.5, and for a diatomic gas, it is 2.5. From the data given in the table, $$\Delta (PV)=-5\ J$$This doesn't seem compatible with the change of -160 J calculated from the work and the heat removed. Thoughts?