The plot of \(V(L)\) vs \(t(^\circ C)\) is shown for an ideal gas. If molecular weight of the gas is 16, how many grams of the gas is used in the experiment? (P = 1 atm)
Show Hint
At \(t = 0^\circ C\), directly use graph value and apply \(PV = nRT\) to avoid unnecessary extrapolation.
Step 1: Understand the graph and physical law.
The graph is a straight line between volume \(V\) and temperature \(t(^\circ C)\), indicating ideal gas behavior. For an ideal gas:
\[
PV = nRT
\]
At constant pressure, volume varies linearly with absolute temperature:
\[
V \propto T
\]
Step 2: Convert Celsius to Kelvin relation.
We use:
\[
T = t + 273
\]
So the graph intercept at \(t = -273^\circ C\) corresponds to \(V = 0\). This confirms linear proportionality between \(V\) and \(T\). Step 3: Use given graph data (key observation).
From the graph, at \(t = 0^\circ C\), volume is approximately:
\[
V_0 = 4.48 \, L
\]
So at \(T = 273 K\), volume is 4.48 L. Step 4: Apply ideal gas equation to find moles.
Using:
\[
PV = nRT
\]
Given:
\[
P = 1 \, \text{atm}, \quad V = 4.48 \, L, \quad T = 273 \, K, \quad R = 0.082
\]
Substitute:
\[
n = \frac{PV}{RT} = \frac{1 \times 4.48}{0.082 \times 273}
\] Step 5: Perform calculation.
First compute denominator:
\[
0.082 \times 273 \approx 22.386
\]
Now:
\[
n = \frac{4.48}{22.386} \approx 0.20 \, \text{mol}
\] Step 6: Convert moles to mass.
Given molar mass = 16 g/mol:
\[
\text{mass} = n \times M = 0.20 \times 16 = 3.2 \, g
\] Final Answer:
\[
\boxed{3.2 \, g}
\]
