34.6g of propane is initially at 0 degrees celsius and 100kPa. If the temp of the gas is accidentally raised to 25 degrees celsius, what mass of propane must be released in order for the volume occupied by the gas to remain constant?

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The answer is 2.9g

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Answer 1 · 2018-02-08T00:49:52

See the full solution below...

First, we must change grams to moles:

moles $C_{3} H_{6} = 34.6 g \div 44 (\frac{g}{mol}) = 0.786$ $C_3H_6 = 34.6 g -: 44 (g/"mol") =0.786$ moles

Now, we are in a position to use the ideal gas law ( also called the "combined gas law"):

$\frac{P_{1} V_{1}}{n_{1} T_{1}} = \frac{P_{2} V_{2}}{n_{2} T_{2}}$ $(P_1V_1)/(n_1T_1)= (P_2V_2)/(n_2T_2)$

The problem seems to imply that both the volume, $V$ $V$ and the pressure $P$ $P$ are to remain constant. This means

$P_{1} = P_{2}$ $P_1=P_2$ and $V_{1} = V_{2}$ $V_1=V_2$

and we cancel these from the equation, leaving

$\frac{1}{n_{1} T_{1}} = \frac{1}{n_{2} T_{2}}$ $(1)/(n_1T_1)= (1)/(n_2T_2)$

or, solving for $n_{2}$ $n_2$

$n_{2} = \frac{n_{1} T_{1}}{T_{2}}$ $n_2 = (n_1T_1)/T_2$

Using the above values (and noting that temperature must be in Kelvin (=Celsius + 273)

$n_{2} = \frac{0.786 \times 273}{298} = 0.720$ $n_2 = (0.786xx273)/298 = 0.720$ moles

If there are to be 0.720 mol of propane remaining in the container, then $0.786 - 0.720 = 0.066$ $0.786 - 0.720 = 0.066$ moles must be released.

At 44.0 g/mol, this mass is 2.88 g (rounded to 2.9 g above).