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Question 9

Half mole of an ideal monoatomic gas is heated at a constant pressure of 1 atm from 20$$^{\circ}$$C to 90$$^{\circ}$$C. Work done by the gas is

Solution

$$\Delta T = T_2 - T_1 = 90 - 20 = 70\text{ K}$$

$$W = P\Delta V$$

$$PV = nRT \implies P\Delta V = nR\Delta T$$

$$W = nR\Delta T$$

$$W = 0.5 \times \frac{25}{3} \times 70 = 35 \times \frac{25}{3} \approx 291\text{ J}$$

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