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Question

(C) The first law of thermodynamics is given as $\Delta U = \Delta Q + \Delta W$. In this convention,$\Delta W$ represents the work done $ON$ the gas.

Vedclass · Accepted Answer

$\Delta U$ is zero when heat is supplied and the temperature stays constant.

Vedclass · Answer

(C) The first law of thermodynamics is given as $\Delta U = \Delta Q + \Delta W$. In this convention,$\Delta W$ represents the work done $ON$ the gas. For option $A$: If no heat enters or leaves,$\Delta Q = 0$,so $\Delta U = \Delta W$. Thus,$\Delta U$ is not necessarily zero. For option $B$: In the equation $\Delta U = \Delta Q + \Delta W$,$\Delta W$ is the work done on the gas,not by the gas. For option $C$: For an ideal gas,internal energy $\Delta U$ is a function of temperature only,i.e.,$\Delta U = nC_v \Delta T$. If the temperature remains constant,$\Delta T = 0$,which implies $\Delta U = 0$. This statement is correct. For option $D$: If temperature increases,$\Delta U 
eq 0$,so $\Delta Q + \Delta W 
eq 0$.

The first law of thermodynamics can be written as $\Delta U = \Delta Q + \Delta W$ for an ideal gas. Which of the following statements is correct?

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When $150 \ J$ of heat is supplied to a system,the work done by the system is $110 \ J$. The change in the internal energy of the system is ....... $J$.

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