Heat and Thermodynamics 6 Question 10
14. In a given process of an ideal gas, $d W=0$ and $d Q<0$. Then for the gas
$(2001$, s)
(a) the temperature will decrease
(b) the volume will increase
(c) the pressure will remain constant
(d) the temperature will increase
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Answer:
Correct Answer: 14. (d)
Solution:
- From first law of thermodynamics,
$$ \begin{array}{llrl} & & d Q & =d U+d W \\ & & d Q & =d U, \text { if } d W=0 \\ \text { Since, } & & d Q & <0 \\ \text { Therefore, } & & d U & <0 \\ \text { or } & U _{\text {final }} & <U _{\text {initial }} \end{array} $$
or temperature will decrease.
NOTE Internal energy $U$ of an ideal gas depends only on the temperature of the gas. Internal energy of $n$ moles of an ideal gas is given by
$$ \begin{aligned} & U=n(f / 2) R T \\ & U \propto T \end{aligned} $$
Here, $f$ is the degree of freedom of the gas.
