Gas heated under constant pressure - thermal energy transferred?

[Marcargo]

Homework Statement

A 1.0 sample of nitrogen gas (diatomic)is heated at constant pressure from 300K to 420K, C^p (nitrogen) = 29.1 Jmol^-1°C^-1

The thermal energy transferred to the gas is?
1. 8370 J
2. 12,200 J
3. 34.90 J
4. 3590 J
5. 12.20 J

Homework Equations

KE = 1/2mv² = 3/2KT (k = 1.38 x 10^-23)
W = PxΔV
γ = 1.4?? (diatomic)
Q = mcΔT?

The Attempt at a Solution

C^-1 (nitrogen) = 29.1 Jmol^-1°C^-1

KE = 1/2mv² = 3/2KT (k = 1.38 x 10^-23)
gives 2.484 x 10 <sup>-21??

Q = mcΔT
= (1 mol x (14 x 2) x 29.1 x (420 - 300)
= 97776 J??

I'm not sure of what other equations to use.
The fact that it is constant pressure suggests that it is isobaric, but I'm not sure how to integrate the formula 'W = PxΔV' with the temperature and the 'moles' of gas.
I think that 'C_p' stands for 'constant pressure' - hence I thought of using Q = MCΔT. I'm not sure whether I'm on the right track, or completely off the bat.

Any help would be appreciated.</sup>

 

[voko]

You are given the molar specific heat at constant pressure, you know that you have one mole, and you have the temperature delta. Sounds very straightforward.

 

[Marcargo]

I'm glad that it's a pretty straight-forward question :).

However, I'm relatively new to all this so it all seems a but confusing.
So, just to be sure, if I have the molar specific heat... what does that mean exactly? That for every 1^°K it's going to gain 29.1°C?

Hmm 29.1°C x (420 - 300) = 3492 J...

Ah.

:)

So just for future reference - If I were to directly convert energy denoted by T^°C into Energy (in Joules), would the formula KE = 0.5mv² = 3/2kT be applicable?

 

[voko]

Molar specific heat (at constant pressure) means that to heat one mole of matter 1 degree Celsius or one kelvin (which is the same) you need to transfer the specified amount of heat (at constant pressure).

To convert energy denoted by temperature in degrees Celsius into joules, you need to convert it to absolute temperature first (in kelvins). This is done by adding 273 K to it. Then you can apply your formula. However, keep in mind that this will give you the energy of one molecule.

 

[Marcargo]

That makes a lot of sense.

Thank You!