Question

2. A diatomic gas with three moles are in a
container at 400 K. Under isobaric process,
its temperature is changed to 900 K. How
much heat is absorbed by the gas during this
process?
​

Answer 1

Answer:

2490J

Explanation:

Let the initial thermodynamic coordinates be $\left(\mathrm{P}_{0}, \mathrm{~V}_{0}, \mathrm{~T}_{0}\right)$, followed by $\left(\mathrm{P}_{1}, \mathrm{~V}_{1}, \mathrm{~T}_{1}\right. )$ and finally$\left(P_{2}, V_{2}, T_{2}\right)$

Given

$\mathrm{T}_{0}=300 \mathrm{~K}$, $\mathbf{P}_{1}=\frac{\mathbf{P}_{0}}{2}$

First process being isochoric,$\mathbf{V}_{1}=\mathbf{V}_{0}$

Thus,$\mathbf{T}_{1}=\frac{\mathbf{T}_{0}}{2}=150 \mathrm{~K}$

Also,$\mathbf{T}_{2}=\mathbf{T}_{0}=2 \mathrm{~T}_{1}=300 \mathrm{~K}$

So, heat released in isochoric cooling= $\Delta \mathrm{Q}_{1}=\mathrm{nC}_{\mathrm{v}} \Delta \mathrm{T}$

and heat absorbed in isobaric heating= $\Delta \mathrm{Q}_{2}=\mathrm{nC}_{\mathrm{p}} \Delta \mathrm{T}$

Netheat absorbed $=\Delta Q_{2}-\Delta Q_{1}=n\left(C_{p}-C_{v}\right) \Delta T=n R \Delta T=2 \times 8.3 \times$150J=2490 J