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The Ideal Gas Equation

If is found, ...

Question

If is found, on heating a gas, its volume increase by 50% and pressure decreases to 60% of its original value. If the original temperature was $- 15^{\circ} C$ , find the temperature to which is was heated?

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Solution

The gas has constant mass as it is ideal. So we can use the equation for the general transformation of the ideal gas:
The mixed gas formula: $\frac{P_{1} V_{1}}{T_{1}} = \frac{P_{2} V_{2}}{T_{2}}$

Given: T1 = -15°C+273 = 258 K

Assume: $P_1$ = P atm and $P_2$ = 60% decraesed, so it has, $P_{2} = \frac{60}{100} \times P$

$V_1$ = V and $V_2$ incraesed by 50%, so it is, $V_{2} = 1 + \frac{50}{100} \times V = \frac{3}{2} \times V$
Plug in the values now,

T2 = $\frac{P 2 V 2}{P 1 V 1}$

T2 =
$\frac{P \times V}{258} = \frac{(\frac{3}{5}) P \times (\frac{3}{2}) V}{T_{2}}$
$T_{2} = \frac{258 \times 3 \times P \times 3 \times V}{5 \times 2 \times P \times V}$
$= 232.2 K$

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