Don't understand this MCAT practice question :/

[CAgirlMx]

---

Members do not see this ad.

At constant pressure, the ratio between the volume expansion of a mole of gas with a high Cp to that of a mole of gas with a low Cp for a given heat input will be

a) greater than 1
b) less than 1
c) equal to 1

equations given:
Q= nCp delta T

equations that come to mind:

since Pressure and moles of gas are constant, Charles Law : V1/T1=V2/T2

Im not understanding why the answer to this question is "b-Less than one"

I would really appreciate the help, thank you!

 

[IOU11]

At constant pressure, the ratio between the volume expansion of a mole of gas with a high Cp to that of a mole of gas with a low Cp for a given heat input will be

a) greater than 1
b) less than 1
c) equal to 1

equations given:
Q= nCp delta T

equations that come to mind:

since Pressure and moles of gas are constant, Charles Law : V1/T1=V2/T2

Im not understanding why the answer to this question is "b-Less than one"

I would really appreciate the help, thank you!

Just rearrange the given equation and solve for n (which I think is moles). you will see that Cp ends up in the denominator. Therefore, If Cp is high, n will be smaller than if Cp was low.. the ratio of moles between the two scenarios would then be less than 1

 

[CAgirlMx]

Just rearrange the given equation and solve for n (which I think is moles). you will see that Cp ends up in the denominator. Therefore, If Cp is high, n will be smaller than if Cp was low.. the ratio of moles between the two scenarios would then be less than 1

I assumed that moles was constant because it says one mole to one mole?

 

[PlsLetMeIn21]

With three answer choices and a topic I cannot find in the AAMC list, I'm not sure you're doing a realistic MCAT question.

 

[mk536]

At constant pressure, the ratio between the volume expansion of a mole of gas with a high Cp to that of a mole of gas with a low Cp for a given heat input will be

a) greater than 1
b) less than 1
c) equal to 1

equations given:
Q= nCp delta T

equations that come to mind:

since Pressure and moles of gas are constant, Charles Law : V1/T1=V2/T2

Im not understanding why the answer to this question is "b-Less than one"

I would really appreciate the help, thank you!

You're on the right track with Charles' law. However you're given heat and not temperature in the problem set up. What you'll want to do is solve for delta T in the given equation:
delta T =Q/(nCp)

Recall that we can also write Charles law as
delta V/delta T= k (*see below if you don't believe me, since this version seems to be provided relatively rarely according to a quick Google search)
Therefore:
(delta V)*n*Cp/Q = k
This means that Cp and delta V are indirectly proportional at a given number of moles and heat input. So a higher Cp results in a lower delta V. Therefore the first gas has a smaller delta V than the second, and the ratio between the two will be less than one.

* from Charles' law
V/T= k
Therefore V(final)=kT(final) and V(initial)=kT(initial).
Subtracting from both sides, we have:
V(final)-V(initial)=kT(final)-kT(initial)
So delta V = k (delta T).