An expansion process on an ideal diatomic ideal gas has a linear path between the initial and final coordinates on a pV diagram.The coordinates of the initial state are: the pressure is 300 kPa,the volume is $0.0700 \mathrm {~m} ^ { 3 }$ and the temperature is $390 \mathrm {~K}$ The final pressure is $130 \mathrm { kPa }$ and the final temperature is $310 \mathrm {~K} .$ How much work is done on the gas during this process? (R = 8.31 J/mol ∙ K)

Question

Quizplus · Accepted Answer

The work done on the gas during an expansion process can be calculated from the area under the process curve on a pV diagram. For a linear process between two states, this area forms a trapezoid. The work done is negative (since it's an expansion, work is done by the gas), and can be calculated using the formula: $W = -\frac{1}{2}(P_{1} + P_{2})(V_{2} - V_{1})$. To find $V_{2}$, use the ideal gas law: $P_{1}V_{1}/T_{1} = P_{2}V_{2}/T_{2}$. Substituting the given values and solving for $V_{2}$ gives $V_{2} = 0.0700 \mathrm{m}^3 \times 300 \mathrm{kPa} / 130 \mathrm{kPa} \times 310 \mathrm{K} / 390 \mathrm{K}$. Then, calculate the work done using the formula for work. The calculation yields a value close to -13,000 J, indicating that approximately 13,000 J of work is done by the gas.

An Expansion Process on an Ideal Diatomic Ideal Gas Has 0.0700 m30.0700 \mathrm {~m} ^ { 3 }0.0700 m3

An Expansion Process on an Ideal Diatomic Ideal Gas Has $0.0700 \mathrm {~m} ^ { 3 }$