Question

The isobar of one mole of an ideal gas was obtained at P atm. The slope of the isobar is $0.82 L K^{-1}$. What is P (in atm)?

• A. 10
• B. 1
• C. 0.1
• D. 0.01

Hint:

Choice of $R$ is crucial. Since the answer is in {atm} and slope involves {Liters}, use $R = 0.0821$. If SI units were used, $R = 8.314$.

Answer 1

The Correct Option is C

Step 1: Understand Isobaric Graph:
An isobar is a graph of Volume ($V$) versus Temperature ($T$) at constant Pressure ($P$). From the ideal gas equation: \[ PV = nRT \] Rearranging for $V$ vs $T$: \[ V = \left( \frac{nR}{P} \right) T \] This represents a straight line equation $y = mx$ passing through origin, where $y=V$, $x=T$, and slope $m = \frac{nR}{P}$.
Step 2: Calculation:
Given: Slope $m = 0.82 \, L K^{-1}$. Moles $n = 1$. Universal Gas Constant $R = 0.0821 \, L \cdot atm \cdot K^{-1} mol^{-1}$ (using atm and L units). Equating slope: \[ \frac{nR}{P} = 0.82 \] \[ \frac{1 \times 0.0821}{P} = 0.82 \] \[ P = \frac{0.0821}{0.82} \] \[ P \approx 0.1 \text{ atm} \]
Step 3: Final Answer:
The pressure is 0.1 atm.