A 12.0 L sample of gas is at STP. What would be its new volume if its pressure was decreased to 575 mmHg and its temperature was doubled?

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A 12.0 L sample of gas is at STP. What would be its new volume if its pressure was decreased to 575 mmHg and its temperature was doubled?

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Answer 1 · 2016-05-05T21:20:26

$V_{2} = 31.7 L$ $V_2=31.7L$

Explanation:

Let us first list the given data:

Position 1:
STP means that $P = 1 a t m = 760 m m H g$ $P=1atm=760mmHg$ and $T = 273 K$ $T=273K$ .
The volume occupied by the gas is $V = 12.0 L$ $V=12.0L$ .

Position 2:
The new conditions are:
$P = 575 m m H g$ $P=575mmHg$ and $T = 2 \times 273 = 546 K$ $T=2xx273=546K$ .
$V = ?$ $V=?$

To find the volume of the gas at position 2, we can rearrange the ideal gas law: $P V = n R T$ $PV=nRT$ to be:

$\frac{P V}{T} = k$ $(PV)/T=k$ where $k = n R$ $k=nR$ , which is constant.

Therefore, $\frac{P_{1} V_{1}}{T_{1}} = \frac{P_{2} V_{2}}{T_{2}}$ $(P_1V_1)/(T_1)=(P_2V_2)/(T_2)$

$=>V_2=(P_1V_1)/(T_1)xx(T_2)/(P_2)=(760cancel(mmHg)xx12.0L)/(273cancel(K))xx(546cancel(K))/(575cancel(mmHg))=31.7L$

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A 12.0 L sample of gas is at STP. What would be its new volume if its pressure was decreased to 575 mmHg and its temperature was doubled?

1 Answer

Explanation:

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