Question

The slope of isobar of one mole of an ideal gas at p (atm) is 0.082 L K$^{-1}$. What is the value of p in atm? (R=0.082 L atm mol$^{-1}$K$^{-1}$)

• A. 0.082
• B. 10
• C. 1
• D. 0.1

Hint:

Ideal gas plots: isobar V ∝ T, slope nR/P. For 1 mole, slope = R/P. Match units of R. Rearrange for P.

Answer 1

The Correct Option is C

1. An isobar is a plot at constant pressure; for ideal gas, $V = n R T / P$.
2. For n=1 mole, $V = (R / P) T$, so slope of V vs T is $R / P$.
3. Given slope = 0.082 L K$^{-1}$, R = 0.082 L atm mol$^{-1}$ K$^{-1}$.
4. Thus, $R / P = 0.082$, so $P = R / 0.082 = 0.082 / 0.082 = 1$ atm.
5. Units confirm: slope in L/K, R/P in (L atm / mol K) / atm = L / mol K, but for n=1, yes.
6. Therefore, the correct option is (3) 1.