Question

0.05 m3 of a gas at 6.9 bar expands reversibly in acylinder behind a piston according to the law
PV1.2 =constant untill volume is 0.08 m3.calculate the work done by the gas and sketch the process on p-vdiagram

Answer 1

Answer on Question #71023, Physics / Molecular Physics | Thermodynamics

$0.05\mathrm{m}^3$ of a gas at 6.9 bar expands reversibly in a cylinder behind a piston according to the law $\mathsf{PV}^{1,2} =$ constant until volume is $0.08\mathrm{m}^3$ . Calculate the work done by the gas and sketch the process on p-v diagram.

Solution:

Work=shaded area

Since,

$$p _ {1} V _ {1} ^ {1. 2} = p _ {2} V _ {2} ^ {1. 2},$$

Therefore

$$p _ {2} = p _ {1} \left(\frac {V _ {1}}{V _ {2}}\right) ^ {1. 2} = 6. 9 \times \left(\frac {0 . 0 5}{0 . 0 8}\right) ^ {1. 2} = 3. 9 2 5 5 9 b a r$$

The work is

$$W = \frac {p _ {2} V _ {2} - p _ {1} V _ {1}}{1 - n}$$ $$W = \frac {3 . 9 2 5 6 \times 0 . 0 8 - 6 . 9 \times 0 . 0 5}{1 - 1 . 2} \times 1 0 ^ {5} = 1 5 4 7 6 J \approx 1 5 4 8 0 J$$

Answer: 15480 J

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