# Question 1b0c9

Mar 14, 2015

You have $\text{0.239 kg}$ of ideal gas in your cylinder.

So, you know the volume your gas occupies, and the temperature and pressure at which it is kept. Because you were given its molar mass, you can solve for its mass by using the ideal gas law equation, $P V = n R T$, to determine how many moles are present in the cylinder.

SInce you know the weight of 1 mole, determining the number of moles will help figure out the weight of the gas.

$P V = n R T \implies n = \frac{P V}{R T}$

n = ((2.14 * 10^(4))/760"atm" * "45.0 L")/(0.082("atm" * "L")/("mol" * "K") * (273.15 + 50)"K") = "47.818 moles of gas"

SIDE NOTE Do not forget to convert mmHg and degrees Celsius to atm and pressure, respectively, if you want to use the value R = 0.082, since it's expressed in atm L/mol K.

This means that its mass in kilograms will be

$\text{47.818 moles" * "5.00 g"/"1 mole" * "1 kg"/"1000 g" = "0.2391 kg}$

Rounded to three sig figs, the answer will be

m_("gas") = "0.239 kg"#