Molar heat capacity

1. May 7, 2015

shrutiphysics

1. The problem statement, all variables and given/known data
A diatomic ideal gas is heated at constant volume until its pressure is doubled. It is again heated at constant pressure until its volume is doubled. The molar heat capacity for the whole process is kR. Find the value of k.

2. Relevant equations
ans is k=19/6.
p/t=constant, v/t=constant(gas laws) . [c][/p]=3.5,[c][/v]=2.5

3. The attempt at a solutioN
let initial temperature be T.Then after 1st step it becomes 2T and after 2nd step it becomes 4T.(gas laws).
Q=2.5*(2T-T)+3.5*(4T-T)
ΔT=3T(is it correct?)

2. May 8, 2015

vela

Staff Emeritus
This doesn't make sense unitwise. Also, you might want to rethink the second $\Delta T$.

3. May 8, 2015

ehild

Q=2.5*(2T-T)+3.5*(4T-T)
ΔT=3T(is it correct?)

Yes, the final temperature is three times the initial one, ΔT=3T is correct. The heat transferred is C n ΔT, but the temperature changes from 2T to 4T in the second step. (And nR is missing from your equation)

4. May 8, 2015

shrutiphysics

Yes, the final temperature is three times the initial one, ΔT=3T is correct. The heat transferred is C n ΔT, but the temperature changes from 2T to 4T in the second step. (And nR is missing from your equation)[/QUOTe
i am not getting the answer in any way.am i using wrong equation???
and how can i find nR???
if anyone knows please guide me to solution as i am stuck badly.

5. May 8, 2015

ehild

What is the heat transferred in the first step?
What is the heat transferred in the second step?

6. May 8, 2015

shrutiphysics

upto my knowledge when i use molar specific heat cp and cv then
heat transferred first step=ncvΔT=2.5*T(temp changes from T to 2T)
2nd step=ncpΔT=3.5*T(temp changes from 2T TO 4T)
And finally to find out molar heat capacity i added both the values and divided it by net temperature change i.e 3T.
where am i going wrong?

7. May 8, 2015

ehild

You said that nCv=2,5 and nCp=3.5. Do you really think that the heat capacities are dimensionless numbers?

8. May 8, 2015

shrutiphysics

Oh sorry i forgot....nCv=2,5R and nCp=3R. R=GAS CONSTANT

9. May 8, 2015

ehild

Cv=2.5R and Cp=3.5R. And Q= C n ΔT.