Question

At 200 K, an ideal gas (X) present in a 1 L flask has a concentration of 1 mol $L^{-1}$. At the same temperature, 0.1 mole of X is added into the vessel. What is the final pressure of the gas in atm? (Given R = 0.082 L atm $mol^{-1} K^{-1}$)

• A. $18.04$
• B. $16.4$
• C. $8.2$
• D. $9.02$

Hint:

When temperature and volume are constant for an ideal gas, the pressure is directly proportional to the number of moles ($P \propto n$).

Answer 1

The Correct Option is A

Step 1: Calculate the initial number of moles ($n_1$).
Concentration $C_1 = 1 \, mol \, L^{-1}$, Volume $V = 1 \, L$.
$n_1 = C_1 \times V = 1 \, mol \, L^{-1} \times 1 \, L = 1 \, mol$.
Step 2: Calculate the initial pressure ($P_1$) using the ideal gas law.
$P_1 V = n_1 R T$ $P_1 = \frac{n_1 R T}{V} = \frac{(1 \, mol) \times (0.082 \, L \, atm \, mol^{-1} \, K^{-1}) \times (200 \, K)}{1 \, L} = 16.4 \, atm$.
Step 3: Calculate the final number of moles ($n_2$).
Moles added $\Delta n = 0.1 \, mol$.
$n_2 = n_1 + \Delta n = 1 \, mol + 0.1 \, mol = 1.1 \, mol$.
Step 4: Calculate the final pressure ($P_2$) using the ideal gas law.
$P_2 V = n_2 R T$ $P_2 = \frac{n_2 R T}{V} = \frac{(1.1 \, mol) \times (0.082 \, L \, atm \, mol^{-1} \, K^{-1}) \times (200 \, K)}{1 \, L} = 1.1 \times 16.4 \, atm = 18.04 \, atm$.
Thus, the final pressure of the gas is $ \boxed{18.04} \, atm $.