Physics -- Change in Internal Energy Help

[MauricioValdez]

Homework Statement

[/B]
Okay guys I have attached a picture of my work.

I guess my question really is, if they are telling me that Cp =7/2 am I allowed to assume that I am dealing with a diatomic gas? If so, that would change my equation to (5/2) instead of (3/2) correct? and therefore my answer would be right.

Homework Equations

[/B]
dU=(3/2)nRdT

The Attempt at a Solution

i have attached my work. My first part is right and second is wrong

 

[ehild]

Homework Statement

[/B]
Okay guys I have attached a picture of my work.

I guess my question really is, if they are telling me that Cp =7/2 am I allowed to assume that I am dealing with a diatomic gas? If so, that would change my equation to (5/2) instead of (3/2) correct? and therefore my answer would be right.

Homework Equations

[/B]
dU=(3/2)nRdT

The Attempt at a Solution

i have attached my work. My first part is right and second is wrong

The internal energy of an ideal two-atomic gas is Cv n T and Cv=Cp-R.

 

[Chestermiller]

Making the determination of whether the gas is diatomic or monoatomic is irrelevant to solving this problem. The first law tells us that:$$\Delta U=Q-W=Q+600$$
For one mole of an ideal gas:$$\Delta U=C_v\Delta T$$and $$\Delta H=C_p\Delta T$$If the process is isobaric, then $$Q=\Delta H$$So, combining these equations, we have$$C_v\Delta T=C_p\Delta T+600$$So, $$(C_p-C_v)\Delta T=-600$$ But, $$C_p-C_v=R$$Therefore, $$R\Delta T=-600$$and $$\Delta U=C_v\Delta T=(C_p-R)\Delta T=\left(\frac{7}{2}-1\right)R\Delta T=-1500$$