The process shown in the p V-diagram is performed on 3.00 moles of a monatomic gas. Determ...

RESEARCH:

From the First law of thermodynamics, heat energy \(Q=W+\Delta E_{\text {int }}\)

Here,

Work done \(=W\)

Change in internal energy \(=\Delta E_{\text {imt }}\)

SIMPLIFY:

Initial volume of the monatomic gas is \(V_{i}=10^{-2} \mathrm{~m}^{3}\)

Final volume of the monatomic gas is \(V_{f}=5.00 \times 10^{-2} \mathrm{~m}^{3}\)

Initial pressure is \(p_{i}=3.00 \times 10^{5} \mathrm{~Pa}\)

Final pressure is \(p_{f}=9.00 \times 10^{5} \mathrm{~Pa}\)

Number of moles of monatomic gas is \(n=3\) moles

CALCULATE:

Equation for work done:

$$ \begin{aligned} W &=\int_{V_{i}}^{V_{f}} P d V \\ &=\frac{1}{2}\left(V_{f}-V_{i}\right)\left(p_{f}-p_{i}\right)+p_{i}\left(V_{f}-V_{i}\right) \\ &=\left(V_{f}-V_{i}\right)\left(\frac{1}{2}\left(p_{f}-p_{i}\right)+p_{i}\right) \\ &=\frac{1}{2}\left(p_{f}+p_{i}\right)\left(V_{f}-V_{i}\right) \end{aligned} $$

Determination of equation for the change in internal energy:

\(T_{f}-T_{i}=\frac{p_{f} V_{f}-p_{i} V_{i}}{n R}\)

Change in internal energy, \(\Delta E=\frac{3}{2} n R\left(\frac{p_{f} V_{f}-p_{i} V_{i}}{n R}\right)\)

\(=\frac{3}{2}\left(p_{f} V_{f}-p_{i} V_{i}\right)\)

Amount of heat input for the process:

$$ \begin{array}{l} Q=W+\Delta E_{\text {int }} \\ =\frac{1}{2}\left[\left(p_{f}+p_{i}\right)\left(V_{f}-V_{i}\right)+3\left(p_{f} V_{f}-p_{i} V_{i}\right)\right] \\ =\frac{1}{2}\left(12.0 \cdot 10^{5} \mathrm{~Pa}\right)\left(4.00 \cdot 10^{-2} \mathrm{~m}^{3}\right)+3\left(9.00 \cdot 10^{5} \mathrm{~Pa}\right)\left(5.00 \cdot 10^{-2} \mathrm{~m}^{3}\right) \\ -\left(3.00 \cdot 10^{5} \mathrm{~Pa}\right)\left(1.00 \cdot 10^{-2} \mathrm{~m}^{3}\right) \\ =8.700 \cdot 10^{4} \mathrm{~J} \end{array} $$

ROUND:

We report our result to three significant figures.

Amount of heat input to this process is \(8.70 \cdot 10^{4} \mathrm{~J}\)