When does raising the temperature of a gas increase its pressure?

Answer 1

Raising the temperature of a gas will only increase the pressure if the volume is conserved, decreased, or (only sometimes) increased. The equation relating all three of these variables is:

PV = nRT, or

Pressure(in pascales) x Volume(in m3) = (number of moles of gas) x 8.3145(J/mol*K) x Temperature(in degrees Kelvin)

We can figure out the answer to your question by using some arbitrary numbers in this equation: we will say that the original Pressure is 101325 pascales (1atm), Volume is 2m3, T is 300oK, and the number of moles of gas is 81.24 moles.

This gives us the equation for the original system:

(101325 pascales)(2m3)=(8.3145J/mol*K)(81.24mol)(300oK)

Let's now increase the temperature by 50oK, to 350oK. We will also conserve the original volume, and see if the pressure increases. This gives us the new equation:

(X pascales)(2m3)=(8.3145J/mol*K)(81.24mol)(350oK).

Since both n and R are constants when considering the same system, we do not need to investigate what will happen when these two are changed (since a change in these two is practically infeasable).

If we solve for Pressure algebraically, we find that:

X pascales = (8.3145J/mol*K)(81.24mol)(350oK)/(2m3) = 118207 pascales.

Since our new pressure, 118207 pascales, is larger than the orgininal pressure, 101325 pascales, we can see that an increase in temperature results in an increase in pressure if volume is conserved.

Additionally, we can investigate what happens when the volume is decreased by 1m3 (from 2m3 to 1m3) along with an increase in temperature by 50oK, to 350oK. We will be comparing this new system to the old system already established in the previous part of this problem. This gives us an equation for the new system:

(X pascales)(1m3)=(8.3145J/mol*K)(81.24mol)(350oK)

If we solve for Pressure algebraically, we find that:

X pascales = (8.3145J/mol*K)(81.24mol)(350oK)/(1m3) = 236414 pascales.

Since our new pressure, 236414 pascales, is larger than the orgininal pressure, 101325 pascales, we can see that an increase in temperature results in an increase in pressure if volume is decreased.

But: what happens if the Volume is increased? Let's consider a situation where an increase in temperature by 50oK, to 350oK is accompanied by a decrease in volume by 1m3, from 2m3 to 3m3. This gives us the equation:

(X pascales)(3m3)=(8.3145J/mol*K)(81.24mol)(350oK)

If we solve for Pressure algebraically, we find that:

X pascales = (8.3145J/mol*K)(81.24mol)(350oK)/(3m3) = 78804 pascales.

Since our new pressure, 78804 pascales, is smaller than the orgininal pressure, 101325 pascales, we can see that an increase in temperature DOES NOT always result in an increase in pressure if volume is increased.

As a general rule, we can therefore conclude that an increase in volume does not always result in an increase in pressure if the temperature is increased. However, there are marginal situations in which this CAN happen: if, for example, the volume is increased by 0.0000001m3, along with an increase in temperature to 350oK, we find that solving for the Pressure gives us a value of 118207 pascales, which is larger than our original pressure, 101325 pascales. So, we can see that an increase in temperature CAN SOMETIMES result in an increase in pressure if volume is increased.

An increase in volume only increases pressure in a system if the increase in volume is significantly small when compared to the increase in temperature.

Answer 2

when volume and the number of particles are constant