Question

A gas is heated at constant volume,what fraction of heat is converted to work​

Answer 1

Given:

A gas is heated at constant volume.

To find:

What fraction of heat is converted to work ?

Calculation:

According to the given question , the amount of work done is an isochoric thermodynamic process (constant volume) will be zero.

$\boxed{ \rm \therefore \: \dfrac{ W}{\Delta Q} = 0}$

So, the fraction of heat used for work done will be zero. This is because the whole heat is used up to increase the internal energy of the system.

Coming to Correct Question:

A gas is heated at constant pressure. What is the fraction of heat used for work done ?

Solution:

According to 1st Law of Thermodynamics:

$\boxed{ \rm \therefore \: \Delta Q = \Delta U + W}$

$\rm \implies \: nC_{p}\Delta \theta = nC_{v}\Delta \theta+ W$

$\rm \implies \: nC_{p}\Delta \theta = n \bigg(\dfrac{C_{p}}{ \gamma } \bigg)\Delta \theta+ W$

$\rm \implies \: W = nC_{p}\Delta \theta \bigg(1 - \dfrac{1}{ \gamma } \bigg)$

$\rm \implies \: W = \Delta Q\bigg(1 - \dfrac{1}{ \gamma } \bigg)$

$\rm \implies \: \dfrac{W}{\Delta Q} = 1 - \dfrac{1}{ \gamma }$

So, the fraction of heat used is :

$\boxed{ \bold{ \: \dfrac{W}{\Delta Q} = 1 - \dfrac{1}{ \gamma } }}$

Answer 2

Answer:

its answer is 0............