Calculate the work done (in calorie) by 10 g of an ideal gas of molar mass 16 g in expanding reversibly and isothermally from a volume of 5 L to 15 L at 300 K.

W =

- 172.48

cal

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W =

- 319.93

cal

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W =

- 352.57.

cal

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W =

- 409.37

cal

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Solution

The correct option is D W = $- 409.37$ cal
Number of moles of ideal gas = $\frac{10}{16}$ mol
R = 1.987 cal $d e g^{- 1} m o l^{- 1}$ T = 300 K
Final volume $V_{2}$ = 15 L , initial volume = 5 L
We know,
W = $- 2.303 n R T l o g \frac{V_{2}}{V_{1}}$
W = $- 2.303 \times \frac{10}{16} \times 1.987 \times 300 \times l o g 3$
W = $- 409.37$ cal

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The pressure-volume work for an ideal gas can be calculated by using the expression $W = \int_{V_{i}}^{V_{f}} p_{e x} d V$ . The work can also be calculated from the pV-plot by using the area under the curve within the specified limits. When an ideal gas is compressed (a) reversibly or (b) irreversibly from volume $V_{i} t o V_{f}$ . Choose the correct option.
(a) W (reversible) = W (irreversible)
(b) W (reversible) < W (irreversible)
(c) W (reversible) = W (irreversible)
(d) W (reversible) = W (irreversible) $+ p_{e x} . Δ V$

2 mole of an ideal gas at $27^{\circ} C$ expands iso - thermally and reversibly from a volume of 4 litre to 40 litre. The work done(in kJ)by the gas is:

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Calculate the work done (in calorie) by 10 g of an ideal gas of molar mass 16 g in expanding reversibly and isothermally from a volume of 5 L to 15 L at 300 K.

The correct option is D W = −409.37 calNumber of moles of ideal gas = 1016 mol R = 1.987 cal deg−1mol−1 T = 300 K Final volume V2 = 15 L , initial volume = 5 L We know, W = −2.303nRT logV2V1 W = −2.303×1016×1.987×300×log 3 W = −409.37 cal