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Standard XIII

Chemistry

Work done in Isothermal Irreversible Process

One mole of a...

Question

One mole of a gas is expanded from ( $1 L, 10 a t m, 300 K$ ) to ( $4 L, 5 a t m, 300 K$ ) against a constant external pressure of $1 a t m$ . The heat capacity of gas is $50 J /^{o} C .$ Then the work done during the process is:
(Take $1 L a t m ≃ 100 J$ )

- 300 J

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- 600 J

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300 J

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- 100 J

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Solution

The correct option is A $- 300 J$
State 1 $\to$ State 2
( $1 L, 10 a t m, 300 K) (4 L, 5 a t m, 300 K$ )

Since, the process is irreversible isothermal expansion (as temperature remains constant) so work done is given as:

$w = - P_{e x t} (V_{2} - V_{1}) = - 1 (4 - 1) = - 3 L a t m = - 3 \times 100 J = - 300 J$

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Work done in Isothermal Irreversible Process

Standard XIII Chemistry

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One mole of a gas is expanded from (1L, 10 atm, 300 K) to (4 L, 5 atm, 300 K) against a constant external pressure of 1 atm. The heat capacity of gas is 50 J/oC.Then the work done during the process is: (Take 1Latm≃100J)

The correct option is A −300 JState 1 → State 2 (1L,10atm,300K)(4L,5atm,300K) Since, the process is irreversible isothermal expansion (as temperature remains constant) so work done is given as: w=−Pext(V2−V1)=−1(4−1)=−3 L atm=−3×100 J=−300 J

One mole of a gas is expanded from ( $1 L, 10 a t m, 300 K$ ) to ( $4 L, 5 a t m, 300 K$ ) against a constant external pressure of $1 a t m$ . The heat capacity of gas is $50 J /^{o} C .$ Then the work done during the process is:
(Take $1 L a t m ≃ 100 J$ )