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When air at 300k and 1 atm pressure is compressed to one.twentieth of its original volume and 30 atm pressure the new temperature will be?

1 Answer

Aug 9, 2018

Using the Gas equation for gases and inserting given values we get

$\frac{P_{1} V_{1}}{T_{1}} = \frac{P_{2} V_{2}}{T_{2}}$ $(P_1V_1)/T_1=(P_2V_2)/T_2$
$\frac{1 \times V_{1}}{300} = \frac{30 \times \frac{V_{1}}{20}}{T_{2}}$ $(1xxV_1)/300=(30xxV_1/20)/T_2$
$\Rightarrow T_{2} = \frac{30 \times \frac{V_{1}}{20}}{\frac{1 \times V_{1}}{300}}$ $=>T_2=(30xxV_1/20)/((1xxV_1)/300)$
$\Rightarrow T_{2} = 30 \times \frac{300}{20}$ $=>T_2=30xx300/20$
$=>T_2=450\ sfK$

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