# Ideal Gas Law Question

#### akbarali

##### New member
The ideal gas law states that the pressure P, volume V , and temperature T of a gas are related by

PV = NkT

where N is the number of molecules of gas, and k is Bolzmann's constant, about 1.38 10^-23 J/K where J is Joules and K is Kelvins. Say that I have 10^24 molecules of gas. The gas begins at a pressure of 200 kPa, inside a 100 cm^3 container, and at a temperature of 400K.

1. Say that I hold the temperature fixed at 400K and begin to decrease the volume of the container at a rate of 10 cm^3/s. At what rate is the pressure changing?

2. What if instead the pressure is kept fixed at 200 kPa, and the gas is cooled at a rate of -2K/s. At what rate is the volume changing? This is what I've got:

Does this seem correct? Comments? Corrections?

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#### MarkFL

Staff member
1.) I agree with your result for the first part, and here's my working:

I begin with:

$$\displaystyle PV=NkT$$

Since $$\displaystyle NkT$$ is constant, differentiating with respect to time $t$ yields:

$$\displaystyle P\frac{dV}{dt}+\frac{dP}{dt}V=0$$

Solving for $$\displaystyle \frac{dP}{dt}$$ there results:

$$\displaystyle \frac{dP}{dt}=-\frac{P}{V}\cdot\frac{dV}{dt}$$

Plugging in the given data, we find:

$$\displaystyle \frac{dP}{dt}=-\frac{200\text{ kPa}}{100\text{ cm}^3}\cdot\left(-10\frac{\text{cm}^3}{s} \right)=20\,\frac{\text{kPa}}{\text{s}}$$

2.) I agree with the method for the second part, but I get a result 1000 times greater:

I begin with:

$$\displaystyle PV=NkT$$

Here $V$ and $T$ are the only changing quantities, so differentiating with respect to time $t$ yields:

$$\displaystyle P\frac{dV}{dt}=Nk\frac{dT}{dt}$$

$$\displaystyle \frac{dV}{dt}=\frac{Nk}{P}\cdot\frac{dT}{dt}$$

Plugging in the given data, we find:

$$\displaystyle \frac{dV}{dt}=\frac{10^{24}\cdot1.38\times10^{-23}\,\frac{\text{J}}{\text{K}}}{200\text{ kPa}}\left(-2\,\frac{\text{K}}{\text{s}} \right)=-138\,\frac{\text{cm}^3}{\text{s}}$$

My answer is 1000 times greater because the standard unit of length in the SI units is the meter (making it $10^6$ greater), and we have kiloPascals reducing it by $10^3$ for a net gain of $10^3$ over your result.

#### akbarali

##### New member
Wow, I did not think to see that. Doh! I'm always missing something! Thankfully, sharp people like yourself can point me in the right direction, hehe.