Question

4.  A 60.0 L tank of chlorine gas at 27 °C and 125 atm springs a leak. When the leak was discovered, the pressure was reduced to 50 atm. How many moles of chlorine gas escaped?   For credit show all work

5   Standard conditions are

a  . 0°C and 14.7 mm
b.    273 K and 760 mm Hg
c     273°C and 760 mm Hg
d     4°C and 7.6 mm Hg

Answer 1

4.

As ,

$PV=nRT$

$n=\frac{PV}{RT}$ , where n= no. of moles

P = pressure , V = volume , R = 0.0821 atm L mol^-1 K^-1 , T = temperature

For calculation of initial moles of chlorine gas,

Given : P= 125 atm

T = 27°C = 273.15 + 27 = 300.15 K

V = 60.0 L

$n_{initial}=\frac{125\;atm \times 60.0\;L}{0.0821\;atm\:L\:mol^{-1}K^{-1}\times 300.15\;K}$

$n_{initial}=304.3545\;mol$

Now , for calculation of final moles of chlorine gas ,

P = 50 atm

T = 27°C = 273.15 + 27 = 300.15 K

V = 60.0L

$n_{final}=\frac{50\;atm\times 60.0\;L}{0.0821\;atm\;L\;mol^{-1}K^{-1}\times 300.15\;K}$

n final 121.7418 mol

Now ,

moles of chlorine escaped = initial moles of chlorine - final moles of chlorine

moles of chlorine escaped = 304.3545 mol - 121.7418 mol

moles of chlorine escaped = 182.6127 mol

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5.

the correct answer is option b that is 273 K and 760 mm Hg