# Question Video: Identifying the Correct Formula Relating Pressure and Temperature of a Gas at a Constant Volume Physics • 9th Grade

Which of the following is the correct formula relating the pressure and temperature of a gas that is kept at a constant volume? š¯‘‡ represents the temperature of the gas, š¯‘¯ represents the pressure of the gas, and š¯‘‰ represents the volume of the gas. [A] š¯‘¯š¯‘‡ = constant [B] š¯‘¯š¯‘‡Ā² = constant [C] š¯‘¯š¯‘‰ = š¯‘‡ [D] š¯‘¯/š¯‘‡ = constant [E] š¯‘¯/š¯‘‡ = š¯‘‰

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### Video Transcript

Which of the following is the correct formula relating the pressure and temperature of a gas that is kept at a constant volume? š¯‘‡ represents the temperature of the gas, š¯‘¯ represents the pressure of the gas, and š¯‘‰ represents the volume of the gas. (A) š¯‘¯ times š¯‘‡ equals a constant. (B) š¯‘¯ times š¯‘‡ squared equals a constant. (C) š¯‘¯ times š¯‘‰ equals š¯‘‡. (D) š¯‘¯ over š¯‘‡ equals a constant. (E) š¯‘¯ over š¯‘‡ equals š¯‘‰.

Alright, so in this question, weā€™re given five different potential formulas that weā€™re told could relate the pressure and temperature of a gas thatā€™s kept at a constant volume. Our task is to work out which of these formulas is the correct one. Now, one necessary condition for a formula to be correct is that the units on the left- and the right-hand side of it must agree with each other. This isnā€™t going to help us with the formulas presented in options (A), (B), and (D) because these all contain a constant on the right-hand side, which could have any kind of units. However, it can help us work out whether option (C) or (E) could be correct.

The three quantities in these formulas are š¯‘¯, the pressure; š¯‘‰, the volume; and š¯‘‡, the temperature of the gas. So letā€™s recall what the units are for each of these quantities. The SI unit for pressure is the pascal, which is equal to the newton per meter squared. Since Newtonā€™s second law states that force is equal to mass multiplied by acceleration, and we can recall that force has SI units of newtons, mass has SI units of kilograms, and acceleration has SI units of meters per second squared, then this means that units of newtons must be equivalent to units of kilograms meters per second squared. Overall then, we have that pascals, the unit for pressure, can be expressed in SI base units as kilograms meters to the minus one seconds to the minus two. Meanwhile, the units for volume are cubic meters, so thatā€™s meters to the power of three. Lastly, temperature is measured in units of kelvin.

So weā€™ve now got the units for each of these three quantities expressed in terms of SI base units. If we consider the units in option (C), then on the left-hand side, thatā€™s units of pressure multiplied by volume. And thatā€™s the same thing as units of pressure multiplied by units of volume. Substituting in our expressions for the units of pressure and volume and then multiplying out the meters to the minus one and the meters cubed, we find that the left-hand side of the equation has units of kilograms meters squared per second squared.

Meanwhile, the units on the right-hand side of the formula are just units of temperature. And we know that the SI base unit for temperature is units of kelvin. Clearly, the units of kelvin on the right-hand side donā€™t agree with the units of kilograms meters squared per second squared on the left-hand side. That means that the formula given in option (C) cannot be correct.

Now, letā€™s consider option (E), which says š¯‘¯ divided by š¯‘‡ is equal to š¯‘‰. Now, we could do the same process that we just did for the formula in option (C). However, we can notice that the right-hand side of the formula, which is just equal to the volume š¯‘‰, must have SI base units of meters cubed. Meanwhile, on the left-hand side, thereā€™s a temperature term in the denominator, which has units of kelvin. And since kelvin donā€™t enter into the expression for the units of pressure, then thereā€™s no way that this kelvin in the denominator can cancel out. So we know that within the units on the left-hand side, thereā€™s a per kelvin, but thereā€™s no kelvin at all in the units on the right. That means that thereā€™s no way for the units on the left and the right to agree. So the formula in option (E) cannot be correct.

At this stage, weā€™ve eliminated two of the potential formulas. And this is as far as we can get by considering the units. To decide between the three remaining available options, we need to consider the physics of the situation. Weā€™re told that weā€™ve got a gas thatā€™s being kept at a constant volume. So we could imagine, for example, that itā€™s being kept inside a box with fixed walls. Since the substance is a gas, then itā€™s made up of a load of particles that are free to move around. And thatā€™s exactly whatā€™s going on. All of the particles are flying around in all different directions inside the box. And as they do this, they can collide with each other and importantly also with the walls of the box.

Supposing that these blue arrows weā€™ve drawn represent the particlesā€™ velocities at some instant in time, then we can see that this particle in the top right is about to hit this right-hand side of the box. Now, since the volume of this box is fixed, then we know that the walls canā€™t move. So when this particle collides with the wall, it will then bounce off the wall and the direction of its velocity will have changed. During this collision, the particle will exert a force on the wall, and thereā€™ll be a component of this force that acts outward.

In this sketch, weā€™ve just drawn a small number of particles to give an idea of whatā€™s going on. But in reality, we would expect there to be a far larger number of particles inside the box. That means that thereā€™s going to be lots of collisions between these particles and the walls of the box. And so thereā€™ll be lots of these forces, each with an outward component, acting on the walls. These outward forces acting over the area of the box walls result in a pressure on the walls of the box.

Now, the faster the particles in the gas are moving, the more force theyā€™re going to exert on the walls when they collide with them. And so the faster the particles, the greater the pressure exerted on the walls. We can recall that the particles in a gas move with some average speed and that this average speed indicates the temperature of the gas. Specifically, the greater this average speed or the faster the particles, the greater the temperature of the gas. Since a greater average speed of the particles means that the gas has a higher temperature and we found that faster-moving particles will result in a greater pressure, then we can combine these two statements to say that higher-temperature particles will exert a greater pressure.

If we look at the formulas given in options (A) and (B), we can see that weā€™ve got pressure times temperature equals a constant or pressure times temperature squared equals a constant. If we have š¯‘¯ multiplied by š¯‘‡ equals a constant, as in option (A), then if the temperature š¯‘‡ increases, the pressure š¯‘¯ must decrease in order for their product to remain constant. But what we found by considering the physics of the situation is that if the temperature š¯‘‡ increases, then the pressure š¯‘¯ must also increase. So then the formula in option (A) cannot be correct.

The same thing is true for option (B), which says that pressure times temperature squared equals a constant. Because again if the temperature š¯‘‡ increases in this formula, then the pressure š¯‘¯ would have to decrease in order for the product š¯‘¯ times š¯‘‡ squared to remain constant. So then we can also eliminate option (B).

This just leaves us with the formula from option (D), which says that š¯‘¯ divided by š¯‘‡ is equal to a constant. Now, we have reasoned that if weā€™ve got a gas kept at a constant volume and we increase its temperature š¯‘‡, then the pressure š¯‘¯ must also increase. The exact relationship between pressure and temperature was discovered back in the 19th century and is known as Gay-Lussacā€™s law. This law says that pressure is directly proportional to temperature, which means that if the temperature of a gas increases, then the pressure will increase in proportion to that. This can equivalently be written as pressure is equal to a constant multiplied by temperature.

We can see that if we take this equation and divide both sides of it by the temperature š¯‘‡, then on the right-hand side, the š¯‘‡ in the numerator cancels with the one in the denominator. That means that Gay-Lussacā€™s law can be written as š¯‘¯ divided by š¯‘‡ is equal to a constant, which is exactly the formula weā€™ve got in option (D). So then this formula from option (D) is our answer to the question. The correct formula relating the pressure and temperature of a gas thatā€™s kept at a constant volume is š¯‘¯ divided by š¯‘‡ is equal to a constant.