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Reversible Isobaric Process
A gas undergoes a reversible isobaric compression. Its initial temperature is $\mathrm{400 K}$ , initial volume is $\mathrm{0.05 m^{3}}$ , and its final volume is $\mathrm{0.02m^{3}}$ . If the pressure remains constant at $\mathrm{ 3 atm}$ during the process, calculate the final temperature.
Option: 1
$\mathrm{160\: k}$

Option: 2
$\mathrm{170\: k}$

Option: 3
$\mathrm{200\: k}$

Option: 4
$\mathrm{20\: k}$

Answers (1)

best_answer

For a reversible isobaric process, the equation relating temperature and volume is:

$\mathrm{ \frac{V_1}{T_1}=\frac{V_2}{T_2} }$

Given:

$\mathrm{ V_1=0.05 \mathrm{~m}^3 }$

$\mathrm{ V_2=0.02 \mathrm{~m}^3 }$

$\mathrm{ T_1=400 \mathrm{~K} }$

Solving for $\mathrm{ T_2 : }$

$\mathrm{T_2 =\frac{V_2}{V_1} \cdot T_1 }$

$\mathrm{T_2 =\frac{0.02 \mathrm{~m}^3}{0.05 \mathrm{~m}^3} \cdot 400 \mathrm{~K} }$

$\mathrm{T_2 =160 \mathrm{~K} }$

Therefore, the correct option is 1.

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