Question #90f5f

2 Answers
May 8, 2017

The final pressure will be #"7.40 atm"#.

Explanation:

The Combined Gas Law can be used to answer this question. It states that the volume of a given amount of gas is proportional to the ratio of its Kelvin temperature and its pressure. The equation is:

#(P_1V_1)/T_1=(P_2V_2)/T_2#

At STP, the standard temperature is #0^@"C"# or #"273.15 K"#. With gas laws, the Kelvin temperature is used. The standard pressure can vary, but since you need the answer in atmospheres, it will be #"1 atm"#. (The more current standard pressure is #"10"^5color(white)(.) "Pa"#, or #"100 kPa"#, or #"1 bar"#, all of which are equal.

To convert #""^@"C"# to Kelvins, add #273.15# to the Celsius temperature.

Organize your data:

Given
#P_1="1 atm"#
#V_1="0.500 L"#
#T_1="273.15 K"#
#V_2=7.50xx10^(-2)color(white)(.)"L"="0.0750 L"#
#T_2="30"^@"C"+273.15="303 K"#

Unknown: #P_2#

Solution
Rearrange the equation to isolate #P_2# and insert the given data into the equation, then solve.

#P_2=(P_1V_1T_2)/(T_1V_2)#

#P_2=(1"atm"xx0.500color(red)cancel(color(black)("L"))xx303color(red)cancel(color(black)("K")))/(273.15color(red)cancel(color(black)("K"))xx0.0750color(red)cancel(color(black)("L")))="7.40 atm"# rounded to three significant figures

May 8, 2017

#"739.7 kPa"#

Explanation:

The gas law states that #PV=nRT#.

In the first part of the question, you are told that the gas occupies #"0.500 L"# at standard temperature and pressure.
this means that conditions are:

  • #"273.15 Kelvin"# (0 celsius)
  • #"100,000 Pascals"# of pressure
  • #"0.500 L"# volume (Which corresponds to #"0.0005 m"^3"#)
    The #R# in the gas law is the Molar gas constant, this value is #"8.314 m"^3 "Pa K"^(−1) "mol"^(−1)"# .

With this information, you can calculate the number of moles of gas you have in your sample.

#PV=nRT#

#n=(PV)/RT#

#n=(100,000*0.0005)/(273.15*8.314)#

#"n=0.02201 Moles"#

With this information, you can now answer the second step. The conditions are now:

  • #"303.15 Kelvin"#
  • #"0.075 L"# (which corresponds to #"0.000075 m"^3"#)
    -#"0.02201 Moles"#
    Now the equation is re-arranged to #P=(nRT)/V#

#P=(0.02201*8.314*303.15)/(0.000075)#

#P="739650.19 Pa"#

The final pressure is #"739650 Pascals"# or #"739.7 kPa"#. This is #"7.30 atm"#.

Hope this helped :D