Heat and Thermodynamics 4 Question 12
14. $2 kg$ of a monoatomic gas is a pressure of $4 \times 10^{4} N / m^{2}$. The density of the gas is $8 kg / m^{3}$. What is the order of energy of the gas due to its thermal motion?
(2019 Main, 10 Jan II)
(a) $10^{6} J$
(b) $10^{3} J$
(c) $10^{4} J$
(d) $10^{5} J$
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Answer:
Correct Answer: 14. (c)
Solution:
- Given, mass of gas, $m=2 kg$
Pressure on gas, $p=4 \times 10^{4} N / m^{2}$
Density of gas, $\quad \rho=8 kg / m^{3}$
$\Rightarrow$ Volume of gas $=\frac{\text { Mass }}{\text { Density }}$
$$ \Rightarrow \quad V=\frac{2}{8}=\frac{1}{4} m^{3} \cdots(i) $$
Using ideal gas equation,
$$ \Rightarrow \begin{aligned} p V & =n R T \\ n R T & =4 \times 10^{4} \times \frac{1}{4} \\ n R T & =10^{4} \cdots(ii) \end{aligned} $$
Internal energy of $n$ moles of a monoatomic gas is given by,
$$ \begin{aligned} U & =\frac{3}{2} n R T \\ \Rightarrow \quad U & =\frac{3}{2} \times 10^{4} J=1.5 \times 10^{4} J \end{aligned} $$
i.e. in order of $10^{4} J$.
