Question

A sample of krypton gas occupies 18.60 l at a pressure of 5.540 atm and 35.00 c.what volume would the krypton occupy at stop

Answer 1

91.33 L

Step-by-step explanation:

For an ideal gas, the product between the gas pressure and the gas volume is proportional to the absolute temperature of the gas. Mathematically:

`pV\propto T`

where:

p is the pressure of the gas

V is the volume of the gas

T is the absolute temperature of the gas

For a transformation of the gas, we can write:

`(p_1 V_1)/(T_1)=(p_2 V_2)/(T_2)`

For the sample of krypton in this problem, we have:

`p_1 = 5.54 atm` (initial pressure)

`V_1 = 18.6 L` (initial volume)

`T_1=35.0^(\circ)+273=308 K` (initial temperature)

`p_2=1 atm` (final pressure, at stp)

`T_2=0^(\circ)+273=273 K` (temperature at stp)

Therefore, the volume at stp is:

`V_2=(p_1 V_1 T_2)/(T_1 p_2)=((5.54)(18.6)(273))/((308)(1))=91.33 L`