 When 1 mole of gas is heated at constant volume, the temperature is raised from 298 to 308 K and heat supplied to gas is 500 J. Which of the following statements is true?A. q = -w = 500J, $\Delta$U=0B. q = $\Delta$U = 500J, w=0C. q=w=500J, $\Delta$U=0D. $\Delta$U=0, q=w= -500J Verified
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Hint: The first law of thermodynamics is the version of the law of conservation of energy, adapted from thermodynamics processes, distinguishing two kinds of transfer of energy as heat and as thermodynamic work and relating them to a function of the body's state called internal energy.
The law of conservation of energy states that the total energy of an isolated system is constant, energy can be transferred from one form to another, but can neither be created nor destroyed.

First law of thermodynamics states that the total energy of an isolated system is constant, energy can be transformed from one form to another but can neither be created nor destroyed.
$\Delta U = q + W$
Where, $\Delta U$represents the change in internal energy of a closed system.
Q denotes the quantity of energy supplied to the system as heat.
W denotes the amount of thermodynamic work done by the system on its surrounding.
An equivalent statement is that a perpetual motion machine of first kind is impossible.
So, from first law of thermodynamics:
$\Delta U = q + W$
and in the given problem, $q = 500J$
$W = - p\Delta V$
where, $\Delta V = 0$
so, $W = 0$
So, $\Delta U = q = 500J$

So, the correct answer is “Option B”.

Note:
In a non-cyclic process, the change in internal energy of the system is equal to net energy added as heat to the system minus the net work done by the system, both being measured in mechanical units.
Taking $\Delta U$ as a change in internal energy one can write,
$\Delta U = q - W$
However, by using IUPAC convention the first law is formulated with work done on the system by its surrounding having a positive sign. With this now often use sign convention for work, the first law for close system can be written as:
$\Delta U = q + W$.