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.

Question

Philoid · Accepted 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 (V₂/V₁)

where V₁ is the initial volume in the state (1) and V₂ is the final volume in the state (2).

For an ideal gas p₁V₁ = p₂V₂

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

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

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