Question

Find the ratio of (DeltaQ)/(DeltaU) and (DeltaQ)/(DeltaW) in an isobaric process. The ratio of molar heat capacities (C_p)/(C_V)=gamma.

Answer 1

$\frac{\Delta Q}{\Delta U}$ = $\gamma$    $\frac{\Delta Q}{\Delta W}$ = $\frac{\gamma -1 }{\gamma }$

Given:

Isobaric process.

The ratio of molar heat capacities $\frac{C_p}{C_V} =\gamma$

To find:

The ratio of ($\frac{\Delta Q}{\Delta U}$) and $\frac{\Delta Q}{\Delta W}$ in an isobaric process.

Formula used:

In an isobaric process, pressure of the system is  = constant

$\Delta U$ = m $C_v$ $\Delta T$

$\Delta Q$ = m $C_p$ $\Delta T$

From first law of thermodynamic.

$\Delta Q$ = $\Delta U$ + $\Delta W$

Explanation:

In an isobaric process, of the system is = constant

$\frac{\Delta Q}{\Delta U}$ = $\frac{mC_p\Delta T}{mC_v\Delta T}$ = $\frac{C_p}{C_v}$ = $\gamma$

From first law of thermodynamic.

$\frac{\Delta Q}{\Delta W}$ = $\frac{\Delta Q}{\Delta Q- \Delta U}$ = 1 - $\frac{\Delta Q}{\Delta U}$ = 1 - $\gamma$

                                   = $\frac{\gamma -1 }{\gamma }$

$\frac{\Delta Q}{\Delta U}$ = $\gamma$    $\frac{\Delta Q}{\Delta W}$ = $\frac{\gamma -1 }{\gamma }$

To learn more....

1)What is isobaric process.

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2)In the isobaric process the ratio of heat supplied to the system and work done by the system for a diaatomic gas is.

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