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Standard XII

Chemistry

Difference between Internal Energy and Enthalpy

1 kg of gas d...

Question

$1 k g$ of gas does $20 k J$ of work and receives $16 k J$ of heat when it is expanded between two states. A second kind of expansion can be found between the initial and final state which requires a heat input of $9 k J$ . The work done by the gas in the second expansion is:

32 k J

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5 k J

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- 4 k J

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13 k J

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Solution

The correct option is D $13 k J$
The change in the internal energy will remain the same, since the initial and final states are the same for both the processes.
Thus, U = Q - W
16 - 20 = 9 - W
$W = 13 k J$

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1 kg

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by absorbing

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Q. A perfect gas goes from a state

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to state

B

by absorbing

8 \times 10^{5} J

and by doing

6.5 \times 10^{5} J

of external work. It is taken from same initial state

A

to final state

B

in another process in which it absorbs

10^{5} J

of heat, then work done in the second process

Q. A perfect gas goes from a state

A

to state

B

by absorbing

8 \times 10^{5} J

and by doing

6.5 \times 10^{5} J

of external work. It is taken from same initial state

A

to final state

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in another process in which it absorbs

10^{5} J

of heat, then work done in the second process

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1 kg of gas does 20 kJ of work and receives 16 kJ of heat when it is expanded between two states. A second kind of expansion can be found between the initial and final state which requires a heat input of 9 kJ. The work done by the gas in the second expansion is:

The correct option is D 13 kJThe change in the internal energy will remain the same, since the initial and final states are the same for both the processes.Thus, U = Q - W16 - 20 = 9 - WW=13 kJ

$1 k g$ of gas does $20 k J$ of work and receives $16 k J$ of heat when it is expanded between two states. A second kind of expansion can be found between the initial and final state which requires a heat input of $9 k J$ . The work done by the gas in the second expansion is:

The correct option is D $13 k J$
The change in the internal energy will remain the same, since the initial and final states are the same for both the processes.
Thus, U = Q - W
16 - 20 = 9 - W
$W = 13 k J$