Question

A sample of propane, a component of lp gas, has a volume of 35.3 l at 315 k and 922 torr. what is its volume at stp (r = 0.08206 l • atm/k • mol, 1 atm = 760 torr)?

Answer 1

To solve this we assume that the gas is an ideal gas. Then, we can use the ideal gas equation which is expressed as PV = nRT. At number of moles the value of PV/T is equal to some constant. At another set of condition of temperature, the constant is still the same. Calculations are as follows:

P1V1/T1 = P2V2/T2

V2 = P1V1T2 / T1P2

V2 = 922(35.3)(273.15) / 315 (760)

V2 = 37.13 L

Answer 2

Answer : The volume of gas at STP is, 37.12 L

Explanation :

Combined gas law is the combination of Boyle's law, Charles's law and Gay-Lussac's law.

The combined gas equation is,

$\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}$

where,

$P_1$ = initial pressure of gas = 922 torr = 1.213 atm

conversion used : (1 atm = 760 torr)

$P_2$ = final pressure of gas at STP = 1 atm

$V_1$ = initial volume of gas = 35.3 L

$V_2$ = final volume of gas at STP  = ?

$T_1$ = initial temperature of gas = 315 K

$T_2$ = final temperature of gas at STP = 273 K

Now put all the given values in the above equation, we get:

$\frac{1.213atm\times 35.3L}{315K}=\frac{1atm\times V_2}{273K}$

$V_2=37.1097L\approx 37.12L$

Therefore, the volume of gas at STP is, 37.12 L