The change in the magnitude of the volume of an ideal gas when a small additional pressure $$\Delta P$$ is applied at a constant temperature, is the same as the change when the temperature is reduced by a small quantity $$\Delta T$$ at constant pressure. The initial temperature and pressure of the gas were 300 K and 2 atm respectively. If $$|\Delta T| = C|\Delta P|$$ then value of C in (K/atm) is __________

For an ideal gas we always start from the ideal‐gas equation

$$PV=nRT.$$

Because the amount of gas does not change, the quantity $$nR$$ is a constant. To find how the volume changes when pressure or temperature change only slightly, we differentiate this equation. The total differential of $$PV=nRT$$ is

$$P\,dV+V\,dP=nR\,dT.$$

Now we divide every term by $$PV$$ so that each differential appears as a fractional (relative) change:

$$\frac{P\,dV}{PV}+\frac{V\,dP}{PV}=\frac{nR\,dT}{PV}.$$

Simplifying each fraction we obtain the useful relation

$$\frac{dV}{V}+\frac{dP}{P}=\frac{dT}{T}.$$

Rearranging this gives

$$\frac{dV}{V}=\frac{dT}{T}-\frac{dP}{P}.$$

This formula tells us how a small change in temperature or pressure affects the volume.

We are told that two different experiments produce equal magnitudes of volume change:

1. A small additional pressure $$\Delta P$$ is applied while the temperature is kept constant (isothermal process).
2. A small decrease in temperature $$\Delta T$$ is made while the pressure is kept constant (isobaric process).

We now evaluate the fractional change in volume for each case.

Isothermal case (constant temperature)

Because the temperature does not change, $$dT=0$$. Substituting $$dT=0$$ into the general differential formula, we have

$$\frac{dV}{V}=0-\frac{dP}{P}=-\frac{dP}{P}.$$

Taking magnitudes (because the statement uses the words “change in magnitude”) gives

$$\left|\frac{\Delta V}{V}\right|=\frac{|\Delta P|}{P}.$$

Isobaric case (constant pressure)

Now the pressure is fixed, so $$dP=0$$. Putting $$dP=0$$ into the same differential relation gives

$$\frac{dV}{V}=\frac{dT}{T}-0=\frac{dT}{T}.$$

Again, taking magnitudes,

$$\left|\frac{\Delta V}{V}\right|=\frac{|\Delta T|}{T}.$$

According to the question these two magnitudes of volume change are equal, so we set them equal:

$$\frac{|\Delta P|}{P}=\frac{|\Delta T|}{T}.$$

The statement $$|\Delta T|=C|\Delta P|$$ is given, so we substitute $$|\Delta T|$$ from the last equation:

$$C|\Delta P| = |\Delta T| = \frac{T}{P}\,|\Delta P|.$$

Because $$|\Delta P|$$ is common on both sides, it cancels, leaving

$$C=\frac{T}{P}.$$

The initial temperature and pressure are provided as $$T=300\ \text{K}$$ and $$P=2\ \text{atm}$$. Substituting these values we get

$$C=\frac{300\ \text{K}}{2\ \text{atm}}=150\ \text{K/atm}.$$

So, the answer is $$150$$.