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:

  1. 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 $10^{4} J$.



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