Question

For an isochoric reversible process, calculate the following if cv = 3R/2 for the gas present in the system and 5 moles of gas is heated from 27oC to 12700.
(a) Work done (W)
(b) Heat supplied (q)
(c) Change in internal energy (AU) ​

Answer 1

Answer:

Heat supplied mark me branliest

Answer 2

Answer:

a) The work done by an ideal gas during an isochoric process is zero.

b)  Heat supplied (q) is calculated by first law of thermodynamics: $q=6235.5 \mathrm{~J}\end{gathered}$

c) Change in internal energy $(\Delta U)$ is given by: $\Delta U=6235.5 J\end{gathered}$

Explanation:

An isochoric process is a process that takes place at constant volume.

The value of ΔV = 0 in an isochoric process.

The work done by an ideal gas during an isochoric process is zero.

Given: If cv = 3R/2 for the gas present in the system and 5 moles of gas is heated from 27oC to 12700.

To find: a) Work done (W) (b) Heat supplied (q) (c) Change in internal energy (AU) ​

Solution:

a)

The work done by an ideal gas during an isochoric process is zero.

(c) Change in internal energy $(\Delta U)$ is given by:

$\begin{gathered}\Delta U=\int_{T_{1}}^{T_{2}} n C_{v} d T \\\Delta U=n C_{v}\left(T_{2}-T_{1}\right) \\\Delta U=5 \times \frac{3 R}{2} \times(400-300) \\\Delta U=750 R \\\Delta U=750 \times 8.314 \\\Delta U=6235.5 J\end{gathered}$

(b) Heat supplied (q) is calculated by first law of thermodynamics:

$\begin{gathered}\Delta U=q+W \\6235.5=q+0 \\q=6235.5 \mathrm{~J}\end{gathered}$

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