Question #cb2ea

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Question #cb2ea

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Answer 1 · 2015-11-13T23:16:31

$2.1 atm$ $"2.1 atm"$

Explanation:

Notice that no mention of volume and number of moles was made, which means that you can assume that they remain constant.

When volume and number of moles are kept constant, pressure and temperature have a direct relationship - this is known as Gay Lussac's Law.

https://elearning.kctcs.edu/bbcswebdav/users/kmuller0001/SoftChalk Filesl

Simply put, if temperature increases, pressure will increase as well. Likewise, if temperature decreases, pressure will decreases as well.

Mathematically, this is written as

$\frac{P_{1}}{T_{1}} = \frac{P_{2}}{T_{2}}$ $color(blue)(P_1/T_1 = P_2/T_2)" "$ , where

$P_{1}$ $P_1$ , $T_{1}$ $T_1$ - the pressure and temperature of the gas at an initial state
$P_{2}$ $P_2$ , $T_{2}$ $T_2$ - the pressure and temperature of the gas at q final state

Now, the temperature of the gas increases from $320 K$ $"320 K"$ to $450 K$ $"450 K"$ . This means that you can expect the pressure of the gas to increases as well.

Plug in your values and solve for $P_{2}$ $P_2$ to get

$P_{2} = \frac{T_{2}}{T_{1}} \cdot P_{1}$ $P_2 = T_2/T_1 * P_1$

$P_{2} = \frac{450 K}{320 K} \cdot 1.5 atm = 2.1 atm$ $P_2 = (450color(red)(cancel(color(black)("K"))))/(320color(red)(cancel(color(black)("K")))) * "1.5 atm" = color(green)("2.1 atm")$

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Question #cb2ea

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