# Homework Help: Thermodynamics - Temperature change of Argon

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1. Dec 3, 2017

### Seven of Nine

1. The problem statement, all variables and given/known data
The temperature of n = 19 mol of argon gas is increased from T1 = 21 oC by Q = 4.4 kJ heat transfer, while the gas pressure is kept constant. What is the new gas temperature in Celsius degrees?

2. Relevant equations
and as its a monoatomic gas I think this means that the equation is just Q=Cp*change in temp. But I'm not really sure. The Cv of Argon is 0.3122, with R being 0.2081 and Cp being 0.5203.

3. The attempt at a solution
My last attempts have come up with the final temp as 44.9 - where I had used the mass of the object as 749. I also have got 30.88 by subbing the values into the unknown. I keep getting the answer marked as wrong and I'm not sure where I'm going wrong

2. Dec 3, 2017

### BvU

HI,

Perhaps you should add dimensions to your quantities and check that the answer has the dimension you want

3. Dec 3, 2017

### Seven of Nine

Do you mean - Cp-0.5203[kJ/kg.K] Cv-0.3122Cp[kJ/kg.K] and then R=0.2081Cp[kJ/kg.K]?

4. Dec 3, 2017

### VSayantan

What is $\Delta T$?

5. Dec 3, 2017

### Seven of Nine

That's what I want in celcius

6. Dec 3, 2017

### VSayantan

Yep. Me too.

But what is it?

Can you elaborate it?

7. Dec 3, 2017

### Seven of Nine

It's the temperature change in celsius.

8. Dec 3, 2017

### VSayantan

cool.

Write that in symbols.

9. Dec 3, 2017

### Staff: Mentor

What is the value of the molar heat capacity at constant volume of a mono atomic gas (in terms of R)? What is the molar heat capacity at constant pressure?

10. Dec 3, 2017

### Seven of Nine

CV = (3/2)R and CP = (5/2)R

Or Q = (3/2)nRΔT + nRΔT = (5/2)nRΔT for Cp

and Q = (3/2)nRΔT for Cv

Is that right?

11. Dec 3, 2017

### Seven of Nine

I have 32.14 when I do - (4400/((5/2)*(759)*0.2081))+21 thats when everything is in grams? Is that correct

12. Dec 3, 2017

### VSayantan

So, $\Delta T$ is change in temperature, right?

Now, say, initially the temperature is $T_i$ and finally $T_f$.
What is the change in temperature?

13. Dec 3, 2017

### Seven of Nine

Yes so the final Temperature was 32.14 - I submitted it and it was correct - with the change in temp being 11.14. This was found but using the molar mass - The molar mass of Argon is 39.95 which is multiplied by the 19mol so 759.05. Then sub into the equation above and then add the initial value to get 32.14

14. Dec 3, 2017

### VSayantan

How did you obtain the value $32.14$?

So your initial equation was wrong!

It should be $$Q=m\cdot {C_p} \cdot {(T_f \sim T_i)}$$

15. Dec 3, 2017

### Seven of Nine

The initial equation is correct I have just subbed in the equation for Cv and the value for n in the initial equation is 1 or excluded as Argon is monoatomic and I got 32.14 by adding the initial value of 21 to the change in Temp which was found using the equation. To give the final Temperature

16. Dec 3, 2017

### VSayantan

The question clearly says "the temperature of $n=19~\rm {mol}$ of argon gas ..." why did you use $n=1$?

How did you find the change in temperature to add with?

17. Dec 3, 2017

### Staff: Mentor

You didn't even need to know the molar mass of argon if you just used the universal gas constant R = 8.314 J/mole.K:
$$(19)(2.5)(8.314)\Delta T = 4400$$
So, $$\Delta T=11.14\ C$$

18. Dec 3, 2017

### VSayantan

Isn't your unit for $T$ - $K$?

19. Dec 3, 2017

### Staff: Mentor

For a temperature change, K and C give the same number.

20. Dec 3, 2017