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A sample of hydrogen at 47°C exerts a pressure of .329 atm. The gas is heated to 77°C at constant volume. What will its new pressure be?

1 Answer

Nov 28, 2015

$P_{2} = 0.360 a t m$ $P_2=0.360atm$

Explanation:

Assuming that Hydrogen is behaving ideally, therefore, we can use the ideal gas law: $P V = n R T$ $PV=nRT$

Since the gas is being heated only and the volume is not changing, therefore, n and V are constant and therefore we can say:

$\frac{P}{T} = k$ $P/T=k$ where $k = \frac{n R}{V}$ $k=(nR)/V$ is a constant value that does not change during the process.

Thus, $\frac{P_{1}}{T_{1}} = \frac{P_{2}}{T_{2}}$ $(P_1)/(T_1)=(P_2)/(T_2)$ This is called Gay-Lussac law.

$\Rightarrow P_{2} = \frac{P_{1}}{T_{1}} \times T_{2}$ $=>P_2=(P_1)/(T_1)xxT_2$

$=>P_2=(0.329atm)/(320cancel(K))xx350cancel(K)=0.360atm$

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