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3 years ago

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5.00 L of a gas is collected at 22.0 'C and 745.0 mmHg . When the temperature increases to 35'C what is the new pressure

1 answer:

Inessa [10]3 years ago

3 0

Answer:

The new pressure is 777.926 mmHg.

Explanation:

We use the ideal gas equation

$PV = nrT$ .

where $P, V , n,$ and $T$ are the pressure, volumes, moles, and temperature of the gas respectively. $r$ is the gas constant $r = 8.3145\: j*mol^{-1}*K^{-1}$ .

Initially, for the gas

$V = 5L = 0.005m^3$

$P = 745.0 \:mmHg = 99325.2\: Pa$

$T = 22^oC = 295\:K.$

therefore, the number of moles $n$ is

$(99325.2\:Pa)(0.005m^3) = n( 8.3145\: j*mol^{-1}*K^{-1})(295K)$

$n = \dfrac{(99325.2\:Pa)(0.005m^3)}{( 8.3145\: j*mol^{-1}*K^{-1})(295K)}$

$n= 0.2025$

Now, when the temperature rises to 35°C, (assuming constant volume) we have

$P = \dfrac{0.2025mol*8.3145\: j*mol^{-1}*K^{-1}308K }{0.005m^3}$

$P = 103715.1 \: Pa$

or in mmHg this is

$\boxed{P = 777.926\:mmHg}$

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