Q. 49

# 1.0 mol of a monoatomic ideal gas is expanded from the state (1) to state (2) as shown in Fig. 6.4. Calculate the work done for the expansion of gas from the state (1) to state (2) at 298 K.

*Fig. :* *6.4*

*Fig. :*

Answer :

o It can be said from Fig. 6.4, that the process of the expansion of the monoatomic ideal gas has been carried out in infinite steps, hence it is obviously an isothermal reversible expansion.

o Hence, the work is done in the process =

W= - 2.303nRT log (V_{2}/V_{1})

where V_{1} is the initial volume in the state (1) and V_{2} is the final volume in the state (2).

For an ideal gas p_{1}V_{1} = p_{2}V_{2}

So (given in the graph)

W= - 2.303nRT log

= - 2.303 × 1 mol × 8.314 J mol^{-1} K^{-1} × 298 K × log2

=- 2.303 × 8.314 × 298 × 0.3010 J

= -1717.46 J

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PREVIOUSEnthalpy diagram for a particular reaction is given in Fig. 6.3. Is it possible to decide the spontaneity of a reaction from the given diagram. Explain.NEXTAn ideal gas is allowed to expand against a constant pressure of 2 bar from 10 L to 50 L in one step. Calculate the amount of work done by the gas. If the same expansion were carried out reversibly, will the work done be higher or lower than the earlier case? (Given that 1 L bar = 100 J)

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