Work done by monatomic gas when 1380 J heat added, temperature rises 272 K to 275 K

[Robershky]

Homework Statement

When 1380 J of heat are added to one mole of an ideal monatomic gas, its temperature increases from 272 K to 275 K. Find the work done by the gas during this process.

Homework Equations

I assume $$\Delta$$U = Q - W is the formula to use. But for some reason I'm just confused.

The Attempt at a Solution

I tried using 1380 = 275 - W, and got -1105. But that is not the correct answer.

 

[Mapes]

Can you relate $\Delta U$ to $\Delta T$?

 

[Robershky]

That's the problem, I don't understand how they relate. Is there a conversion from Kelvin to Joules? Or do I just use 3K = 1380J - W?

 

[Mapes]

Check your book, notes, or reference for an equation of the internal energy (U) of an ideal gas.

 

[Brian_C]

Internal energy increases with increasing temperature, but the exact relation will depend on the type of gas. You should figure out how many degrees of freedom are present in a monatomic gas, then use equipartition of energy to express the internal energy as a function of temperature.

 

[Andrew Mason]

Homework Statement

When 1380 J of heat are added to one mole of an ideal monatomic gas, its temperature increases from 272 K to 275 K. Find the work done by the gas during this process.

Homework Equations

I assume $$\Delta$$U = Q - W is the formula to use. But for some reason I'm just confused.

The Attempt at a Solution

I tried using 1380 = 275 - W, and got -1105. But that is not the correct answer.

You are given Q and you can determine $\Delta U = nC_v\Delta T$. So it is just a matter of applying the first law to determine W.

AM