Molar Heat Capacity: Find Value of k

[shrutiphysics]

Homework Statement

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.

Homework 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?)
this is all i can think.please help.i am not getting answer.

 

[vela]

Homework Statement

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.

Homework 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)

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

ΔT=3T(is it correct?)
this is all i can think.please help.i am not getting answer.

 

[ehild]

Homework Statement

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.

Homework 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?)
this is all i can think.please help.i am not getting answer.[/QUOTE]

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)

 

[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

Q=2.5*(2T-T)+3.5*(4T-T)
ΔT=3T(is it correct?)
this is all i can think.please help.i am not getting answer.

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.

 

[ehild]

Show your work please.
What is the heat transferred in the first step?
What is the heat transferred in the second step?

 

[shrutiphysics]

Show your work please.
What is the heat transferred in the first step?
What is the heat transferred in the second step?

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?

 

[ehild]

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

 

[shrutiphysics]

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

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

 

[ehild]

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

Cv=2.5R and Cp=3.5R. And Q= C n ΔT.