Question

A diatomic gas does 200 j of work when it expands isobarically. Find the heat given to the gas in the process​ !!​

Answer 1

The work done by the gas, $\sf{\Delta W=200\ J.}$

As the process is isobaric, $\sf{\Delta W=P\Delta V.}$

By First Law of Thermodynamics, the heat supplied is,

$\sf{\longrightarrow \Delta Q=\Delta u+\Delta W}$

$\sf{\longrightarrow \Delta Q=\dfrac{f}{2}\,nR\Delta T+P\Delta V}$

Since $\sf{P\Delta V=nR\Delta T,}$

$\sf{\longrightarrow \Delta Q=\dfrac{f}{2}\,P\Delta V+P\Delta V}$

$\sf{\longrightarrow \Delta Q=\left(\dfrac{f}{2}+1\right)\,P\Delta V}$

Here $\sf{f}$ is the degree of freedom.

Since the gas is diatomic, $\sf{f=5.}$

Then,

$\sf{\longrightarrow \Delta Q=\left(\dfrac{5}{2}+1\right)200}$

$\sf{\longrightarrow\underline{\underline{\Delta Q=700\ J}}}$

Answer 2

Explanation:

Given :

• A diatomic gas does 200 j of work when it expands isobarically.

To Find :

• Find the heat given to the gas in the process !!

Solution :

We Know ,

The ratio of the $\sf\dfrac{C_{p}}{C_{v}}$

For diatomic gas is$\sf\dfrac{C_{p}}{C_{v}}= \dfrac{7}{5}$

According to the isobaric process is :

• An Isobaric process is a thermodynamic process in which the pressure stays constant: ΔP = 0.

• The heat transferred to the system does work, but also changes the internal energy of the system.

• Using this convention, by the first law of thermodynamics, where W is work, U is internal energy, and Q is heat.

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$\sf \implies W = p(V_2 -V_1) \\ \\\sf \implies W = nRT_2 - nRT_1 \\ \\\sf \implies T_2 - T_1 = \frac{W}{nR}$

The heat given in an isobaric process is

$\sf \implies Q = nC_p(T_2 - T_1) \\ \\\sf \implies Q = nC_p \frac{W}{nR} = \frac{7}{2}W \\ \\\sf \implies Q = \frac{7}{2} \times 200 \\ \\\sf \implies Q = 700\:J$

• Hence the Q is 700 j