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प्रश्न
1.0 mol of a monoatomic ideal gas is expanded from state (1) to state (2) as shown in figure. Calculate the work done for the expansion of gas from state (1) to state (2) at 298 K.
उत्तर
It is clear from the figure that the process has been carried out in infinite steps, hence it is isothermal reversible expansion.
w = – 2.303 nRT log `V_2/V_1`
But, p1V1 = p2V2
⇒ `V_2/V_1 = p_1/p_2 = 2/1` = 2
∴ w = – 2.303 nRT log `p_1/p_2`
= – 2.303 × 1 mol × 8.314 J mol–1 K–1 × 298 K–1 × log 2
= – 2.303 × 8.314 × 298 × 0.3010 J
= –1717.46 J
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संबंधित प्रश्न
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Match the following :
| A | B |
| (i) Adiabatic process | (a) Heat |
| (ii) Isolated system | (b) At constant volume |
| (iii) Isothermal change | (c) First law of thermodynamics |
| (iv) Path function | (d) No exchange of energy and matter |
| (v) State function | (e) No transfer of heat |
| (vi) ΔU = q | (f) Constant temperature |
| (vii) Law of conservation of energy | (g) Internal energy |
| (viii) Reversible process | (h) Pext = o |
| (ix) Free expansion | (i) At constant pressure |
| (x) ΔH = q | (j) Infinitely slow process which proceeds through a series of equilibrium states. |
| (xi) Intensive property | (k) Entropy |
| (xii) Extensive property | (l) Pressure |
| (m) Specific heat |
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