When the pressure of a gas is halved and its temperature is doubled, what happens to its volume?
1 Answer
The volume of the gas will increase by a factor of
Explanation:
Your tool of choice here will be the combined gas law equation, which looks like this
color(blue)(ul(color(black)((P_1V_1)/T_1 = (P_2V_2)/T_2)))
Here
P_1 ,V_1 ,T_1 are the pressure, volume, and absolute temperature of the gas at an initial stateP_2 ,V_2 ,T_2 are the pressure, volume, and absolute temperature of the gas at a final state
Now, notice that when the temperature is kept constant, increasing the pressure of the gas by a specific factor will cause its volume to decrease by the same factor
Similarly, when the pressure is kept constant, increasing the temperature of the gas by a specific factor will cause its volume to increase by the same factor
This tells you that increasing the temperature of the gas will cause its volume to increase. Similarly, decreasing its pressure will also cause its volume to increase.
So even without doing any calculations, you should be able to say that
V_2 > V_1
In other words, decreasing the pressure of the gas and increasing its temperature are changes that do not compete with each other in terms of their influence on the volume of the gas.
In your case, you have
P_2 = P_1/2 -> the pressure of the gas is halved
T_2 = 2 * T_1 -> the temperature of the gas is doubled
Rearrange the combined gas law equation to solve for
V_2 = P_1/P_2 * T_2/T_1 * V_1
Plug in your values to find
V_2 = overbrace(color(red)(cancel(color(black)(P_1)))/(color(red)(cancel(color(black)(P_1)))/2))^(color(blue)("influence of pressure")) * overbrace( (2 * color(red)(cancel(color(black)(T_1))))/color(red)(cancel(color(black)(T_1))))^(color(blue)("influence of temperature")) * V_1
V_2 = 2 * 2 * V_1
color(darkgreen)(ul(color(black)(V_2 = 4 * V_1)))
As predicted, the volume of the gas increased as a result of the two changes. Notice that the volume increased by a factor that is equal to the product of the factor that corresponds to the decrease in pressure, i.e.