Question #d400a

We can solve this problem using the pressure- volume relationship of gases, illustrated by Boyle's law:

#P_1V_1 = P_2V_2#

We can see that pressure and volume for gases are inversely proportional, indicating that as one increases, the other must decrease (assuming constant temperature). Since the volume decreased, we can expect an increase in the pressure of the gas.

To keep calculations consistent, let's convert the units #"dm"^3# to #"m"^3# in the final volume measurement:

#10.0 cancel("dm"^3)((1 "m"^3)/((10.0)^3 cancel("dm"^3))) = 0.0100 "m"^3#

Now that we have our necessary variables, let's rearrange the Boyle's law equation to solve for the final pressure, #P_2#:

#P_2 = (P_1V_1)/(V_2)#

Now, plugging in known variables, we have

#P_2 = ((100,000"N"/("m"^2))(50.0 cancel("m"^3)))/((0.0100 cancel("m"^3))) = color(red)(5.00 xx 10^8 "N"/("m"^2)#

The units for pressure above, #"N"/("m"^2)#, is equivalent to a unit of pressure called the pascal (symbol #"Pa"#):

#1 "N"/("m"^2) = 1 "Pa"#

Thus, the pressure can also be reported as #color(red)(5.00 xx 10^8 "Pa"#

If you want another unit of pressure, a more common one when dealing with gases, we can convert this to atmospheres using the conversion factor #(1 "atm")/(101,325 "Pa")#:

#5.00 xx 10^8 cancel("Pa")((1 "atm")/(101,325 cancel("Pa"))) = color(blue)(4.93 xx 10^3 "atm"#