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Question: A thermodynamic process, \[{\text{200}}\,{\text{J}}\] of heat is given to a gas and \[{\text{100}}\,...

A thermodynamic process, 200 J{\text{200}}\,{\text{J}} of heat is given to a gas and 100 J{\text{100}}\,{\text{J}} of work is also done on it. The change in internal energy of the gas is:
A. 100 J100\,{\text{J}}
B. 300 J300\,{\text{J}}
C. 419 J419\,{\text{J}}
D. 24 J24\,{\text{J}}

Explanation

Solution

Change in internal energy is related to the heat given and work done on the system or by the system. To find the change in internal energy use the first law of thermodynamics. It actually represents the law of conservation of energy.

Complete Step by step answer: Given, heat given to the system, ΔQ=200 J\Delta Q = 200\,{\text{J}}
While putting the values of the work done and heat added we should keep in mind that whether work is done on the system or by the system and take the sign accordingly.
And work is done on the system, so ΔW=−100 J\Delta W = - 100\,{\text{J}}
Let ΔU\Delta U be the change in internal energy.
From first law of thermodynamic we have,

ΔU=ΔQ−ΔW\Delta U = \Delta Q - \Delta W

Now, putting the values of ΔQ\Delta Q and ΔW\Delta W in the above equation, we get

\Rightarrow \Delta U = 300\,{\text{J}} \\\ $$ Therefore, change in internal energy is $$300\,{\text{J}}$$ **Hence, the correct answer is option (B) $$300\,{\text{J}}$$** **Note:** There are some important points we need to keep in mind in thermodynamics problems. Always remember heat given to a system is taken to be positive and heat given by the system is taken to be negative. Similarly, work done by the system is taken to be positive and work done on the system is taken to be negative.