The gas inside of a container exerts #8 Pa# of pressure and is at a temperature of #220 ^o K#. If the temperature of the gas changes to #20 ^oC# with no change in the container's volume, what is the new pressure of the gas?

Question

The gas inside of a container exerts #8   Pa# of pressure and is at a temperature of #220 ^o K#.  If the temperature of the gas changes to #20 ^oC# with no change in the container's volume, what is the new pressure of the gas?

Benjamin Asiedu · Accepted Answer

The final pressure is #10.66# Pa.

I assume that we are working with a perfect gas. The process is called isochoric (it means that mantains the volume unchanged). Those process follow the law #P/T=#constant. This mean that if we have an initial pressure and temperature #P_i, T_i# and a final pressure and temperature #P_f, T_f# the law tell us: #P_i/T_i=#constant #P_f/T_f=#constant then #P_i/T_i=P_f/T_f#. In our case we have #P_i=8# Pa #T_i=220# K (you shouldn't use the symbol #\circ# for Kelvin) #T_f=20^\circ# C and we want to know #P_f#. It is always better to work in the International System then we convert the Celsius in Kelvin adding #273.15#: #T_f=20^\circ# C #=293.15# K. Before solving the equation of the isochoric process we can notice that the temperature is increasing of 73.15 K from 220 K to 293.15 K. If we warm up a gas we expect that the pressure increase because the particles of the gas move faster. So at the end we should find something bigger than 8 Pa. We can write our equation for isochoric process #8/220 = P_f/293.15# obtaining #P_f=8/220 293.15=10.66# Pa (that is bigger than 8, as expected from a warmer gas).

William Audy · Answer

undefined To find the new pressure of the gas when the temperature changes, you can use the ideal gas law equation:

$$ P_1 \times V_1 / T_1 = P_2 \times V_2 / T_2 $$

Given that the volume and volume remain constant, we can simplify the equation to:

$$ P_1 / T_1 = P_2 / T_2 $$

Substituting the given values:

$$ 8 \, \text{Pa} / 220 \, \text{K} = P_2 / (20 + 273.15) \, \text{K} $$

Solving for $ P_2 $:

$$ P_2 = (8 \, \text{Pa} \times (20 + 273.15) \, \text{K}) / 220 \, \text{K} $$

$$ P_2 ≈ 72.36 \, \text{Pa} $$

Therefore, the new pressure of the gas is approximately 72.36 Pa.