61.5L of oxygen at 18°C and an absolute pressure of 2.45 atm is compressed to 48.8L and at the same time the temperature is raised to 50°C . what is the new pressure of the gas?
1 Answer
Explanation:
Notice that you're dealing with a clear-cut case of a combined gas law problem.
More specifically, you know that pressure, temperature, and volume are all changing from the initial state of the gas to the final state of the gas.
This of course in the context of the number of moles of gas being kept constant.
So let's say that you're not familiar with the combined gas law equation. Use the ideal gas law equation to describe the gas at the initial state
#P_1 * V_1 = n * R * T_1#
and at the final state
#P_2 * V_2 = n * R * T_2#
Divide these two equations to get
#(P_1V_1)/(P_2V_2) = (color(red)(cancel(color(black)(n * R))) * T_1)/(color(red)(cancel(color(black)(n * R))) * T_2)#
This is equivalent to
#color(blue)( (P_1V_1)/T_1 = (P_2V_2)/T_2) -># the combined gas law equation
It is very important to remember that the temperature of the gas must be expressed in Kelvin!
You want to know the final pressure of the gas, so rearrange this equation and solve for
#P_2 = V_1/V_2 * T_2/T_1 * P_1#
Plug in your values to get
#P_2 = (61.5color(red)(cancel(color(black)("L"))))/(48.8color(red)(cancel(color(black)("L")))) * ((273.15 + 50)color(red)(cancel(color(black)("K"))))/((273.15 + 18)color(red)(cancel(color(black)("K")))) * "2.45 atm"#
#P_2 = "3.4270 atm"#
Now, you should round this off to one sig fig, the number of sig figs you have for the second temperature of the gas, but I'll leave it rounded off to two sig figs
#P_2 = color(green)("3.4 atm")#
