Is combined gas law only when moles are same

 Is combined gas law only when moles are the same?

Yes, the Combined Gas Law assumes that the number of moles ($n$) of the gas remains constant. It combines three fundamental gas laws — Boyle's Law, Charles's Law, and Gay-Lussac's Law — into a single relationship:

$$\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}$$

Where:

• $P_1$, $P_2$ = initial and final pressures
• $V_1$, $V_2$ = initial and final volumes
• $T_1$, $T_2$ = initial and final temperatures (in Kelvin)

For this equation to hold, $n$ (the number of moles) must remain constant. If the moles of gas change, the equation would need to incorporate the Ideal Gas Law:

$$PV = nRT$$

In that case, you would need to account for changes in $n$ explicitly.

The Combined Gas Law:

$$\frac{P_1{V}_1}{T_1}=\frac{P_2{V}_2}{T_{\mathrm{2<span\; class="vlist"><span\; class="mord"><span\; class="mord"><span\; class="msupsub"><span\; class="vlist-t\; vlist-t2"><span\; class="vlist-r"><span\; class="vlist-s"></span></span><span\; class="vlist-r"><span\; class="vlist"></span></span></span></span></span></span></span><span\; class="vlist-s"></span>}}}$$

This relationship is valid only if the number of moles of gas remains constant.

Example 1: Inflating a Balloon (Moles Constant)

Suppose you inflate a balloon at sea level where the pressure is $P_1 = 1 \, \text{atm}$, the temperature is $T_1 = 300 \, \text{K}$, and the volume is $V_1 = 2 \, \text{L}$.

If the balloon rises to a high altitude where the pressure drops to $P_2 = 0.5 \, \text{atm}$ and the temperature decreases to $T_2 = 250 \, \text{K}$, what is the new volume of the balloon?

Solution:
Using the combined gas law:

$$\frac{P_1{V}_1}{T_1}=\frac{P_2{V}_2}{T_{\mathrm{2<span\; class="vlist"><span\; class="mord"><span\; class="mord"><span\; class="msupsub"><span\; class="vlist-t\; vlist-t2"><span\; class="vlist-r"><span\; class="vlist-s"></span></span><span\; class="vlist-r"><span\; class="vlist"></span></span></span></span></span></span></span><span\; class="vlist-s"></span>}}}$$

Substitute known values:

$$\frac{(1)(2)}{300} = \frac{(0.5)(V_2)}{250}$$

Solve for $V_2$:

$$V_2 = \frac{(1)(2)(250)}{(300)(0.5)} = 3.33 \, \text{L}$$

So, the balloon’s volume increases to $3.33 \, \text{L}$.

Example 2: Changing Moles (Ideal Gas Law)

Suppose $2 \, \text{mol}$ of gas is in a cylinder at $P = 2 \, \text{atm}$, $V = 10 \, \text{L}$, and $T = 300 \, \text{K}$. Now, an additional $1 \, \text{mol}$ of gas is added, making $n = 3 \, \text{mol}$. What is the new pressure if the temperature and volume remain constant?

Solution:
Use the Ideal Gas Law:

$$PV = nRT$$

For initial conditions:

$$P_1 = \frac{n_1 R T}{V} = \frac{(2)(R)(300)}{10} = 60R$$

For final conditions:

$$P_2 = \frac{n_2 R T}{V} = \frac{(3)(R)(300)}{10} = 90R$$

Thus:

$$P_2 = \frac{90R}{60R} \cdot P_1 = 1.5 \, \text{times the original pressure}$$
$$P_2 = 3 \, \text{atm}$$

This shows that when moles change, the combined gas law is insufficient, and the Ideal Gas Law must be used.