In a thermodynamic process,the pressure of a | vedclass

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(B) According to the first law of thermodynamics,$\Delta Q = \Delta U + \Delta W$.Given that the gas releases $20 \, J$ of heat,the heat supplied is $

Vedclass · Accepted Answer

$18$

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(B) According to the first law of thermodynamics,$\Delta Q = \Delta U + \Delta W$.Given that the gas releases $20 \, J$ of heat,the heat supplied is $\Delta Q = -20 \, J$.Since $8 \, J$ of work is done on the gas,the work done by the gas is $\Delta W = -8 \, J$.Substituting these values into the first law equation:$-20 \, J = \Delta U + (-8 \, J)$.$\Delta U = -20 \, J + 8 \, J = -12 \, J$.We know that $\Delta U = U_f - U_i$,where $U_f$ is the final internal energy and $U_i$ is the initial internal energy.Given $U_i = 30 \, J$,we have:$U_f - 30 \, J = -12 \, J$.$U_f = 30 \, J - 12 \, J = 18 \, J$.Therefore,the final internal energy is $18 \, J$.

In a thermodynamic process,the pressure of a fixed mass of a gas is changed in such a manner that the gas releases $20 \, J$ of heat when $8 \, J$ of work is done on the gas. If the initial internal energy of the gas was $30 \, J$,then the final internal energy will be .... $J$.

Similar Questions

The first law of thermodynamics states that:

The volume of one mole of the gas is changed from $V$ to $2V$ at constant pressure $p$. If $\gamma$ is the ratio of specific heats of the gas,what is the change in internal energy of the gas?

If the heat given to a system is $35 \ J$ and the work done on the system is $15 \ J$,then the change in the internal energy of the system is ... $J$?

If $\Delta Q$ and $\Delta W$ represent the heat supplied to the system and the work done on the system respectively,then the first law of thermodynamics can be written as,where $\Delta U$ is the change in internal energy.

$Assertion:$ The heat supplied to a system is always equal to the increase in its internal energy.
$Reason:$ When a system changes from one thermal equilibrium to another,some heat is absorbed by it.

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