Thermodynamics - Temperature change of Argon

[Seven of Nine]

Homework Statement

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?

Homework 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.

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

 

[BvU]

HI,

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

 

[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]?

 

[VSayantan]

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

What is $\Delta T$?

 

[Seven of Nine]

That's what I want in celcius

 

[VSayantan]

That's what I want in celcius

Yep. Me too.

But what is it?

Can you elaborate it?

 

[Seven of Nine]

It's the temperature change in celsius.

 

[VSayantan]

It's the temperature change in celsius.

cool.

Write that in symbols.

 

[Chestermiller]

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?

 

[Seven of Nine]

What is the molar heat capacity at constant pressure?

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?

 

[Seven of Nine]

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

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

 

[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?

 

[Seven of Nine]

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?

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

 

[VSayantan]

Then sub into the equation above and then add the initial value to get 32.14

How did you obtain the value $32.14$?

So your initial equation was wrong!

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

 

[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

 

[VSayantan]

the value for n in the initial equation is 1

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

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

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

 

[Chestermiller]

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

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, $$\Delta T=11.14\ C$$

 

[VSayantan]

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, $$\Delta T=11.14\ C$$

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

 

[Chestermiller]

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

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

 

[VSayantan]

Yes.

You're absolutely right @Chestermiller