Question 83: Thermodynamics & KTG

Question

In mm$$^3$$ of a gas is compressed at 1 atmospheric pressure and temperature 27$$^\circ$$C to 627$$^\circ$$C. What is the final pressure under adiabatic condition?
($$\gamma$$ for the gas = 1.5)

Accepted Answer

Given,
$$\begin{aligned}
V_1 & =1 \mathrm{~mm}^3 \
& =10^{-9} \mathrm{~m}^3 \
p_1 & =1 \mathrm{~atm} \
& =1 	imes 10^5 \mathrm{~N} / \mathrm{m}^2 \
T_1 & =27+273 \
& =300 \mathrm{~K} \
T_2 & =627+273 \
& =900 \mathrm{~K}
\end{aligned}$$
Under adiabatic condition,
$$\begin{aligned}
p_1^{1-\gamma} T_1^\gamma & =p_2^{1-\gamma} T_2^\gamma \
\Rightarrow \quad\left(\frac{p_2}{p_1}ight)^{1-\gamma} & =\left(\frac{T_1}{T_2}ight)^\gamma \
\Rightarrow \quad\left(\frac{p_2}{p_1}ight)^{1-1.5} & =\left(\frac{300}{900}ight)^{1.5} \
\Rightarrow \quad\left(\frac{p_2}{p_1}ight)^{-0.5} & =\left(\frac{1}{3}ight)^{1.5} \
\Rightarrow \quad\left(\frac{p_1}{p_2}ight)^{0.5} & =\left(\frac{1}{3}ight)^{1.5}
\end{aligned}$$
$$\begin{aligned} & \Rightarrow \quad \frac{p_1}{p_2}=\left(\frac{1}{3}ight)^{\frac{1.5}{0.5}}=\left(\frac{1}{3}ight)^3=\frac{1}{27} \ & \Rightarrow \quad p_2=27 p_1=27 	imes 1 	imes 10^5 \ & =27 	imes 10^5 \mathrm{~N} / \mathrm{m}^2 \ & \end{aligned}$$

Answer

$$27	imes10^5$$ N/m$$^2$$

Answer

$$80	imes10^5$$ N/m$$^2$$

Answer

$$36	imes10^5$$ N/m$$^2$$

Answer

$$56	imes10^5$$ N/m$$^2$$

Thermodynamics & KTG

In mm$^3 of a gas is compressed at 1 atmospheric pressure and temperature 27^\circC to 627^\circC. What is the final pressure under adiabatic condition? (\gamma$ for the gas = 1.5)

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