Question 5: Thermodynamics & KTG

Question

An ideal gas initially at pressure 1 bar is being compressed from $$30 \mathrm{~m}^3$$ to $$10 \mathrm{~m}^3$$ volume and its temperature decreases from $$320 \mathrm{~K}$$ to $$280 \mathrm{~K}$$, then find final pressure of the gas.

Accepted Answer

Given, initial pressure of ideal gas,
$$p_1=1 	ext { bar }=1 \mathrm{~N} / \mathrm{m}^2$$
Initital volume $$V_1=30 \mathrm{~m}^3$$
Final volume, $$V_2=10 \mathrm{~m}^3$$
Initial temperature, $$T_1=320 \mathrm{~K}$$
Final temperature, $$T_2=280 \mathrm{~K}$$
Final pressure of gas, $$p_2=$$ ?
By ideal gas equation,
$$\begin{aligned}
& \frac{p_1 V_1}{T_1}=\frac{p_2 V_2}{T_2} \Rightarrow \frac{1 	imes 30}{320}=\frac{p_2 	imes 10}{280} \
\Rightarrow \quad & p_2=2.625 \mathrm{~bar}
\end{aligned}$$

Answer

2.625 bar

Answer

3.4 bar

Answer

1.325 bar

Answer

4.5 bar

Thermodynamics & KTG

An ideal gas initially at pressure 1 bar is being compressed from $30 \mathrm{~m}^3 to 10 \mathrm{~m}^3 volume and its temperature decreases from 320 \mathrm{~K} to 280 \mathrm{~K}$, then find final pressure of the gas.

Choose the correct answer: