Question

A cylinder contains 18 moles of oxygen at pressure of 15 atmosphere at temperature 300 K. If the pressure reduces to 9 atmospheres by the withdrawal of 6 moles of oxygen, then the temperature of the cylinder will be reduced to

• A. 200 K
• B. 230 K
• C. 270 K
• D. 220 K
• E. 250 K

Hint:

Logic Tip: By observing ratios, you can solve this mentally. The pressure ratio $P_2/P_1$ is $9/15 = 3/5$. The mole ratio $n_2/n_1$ is $12/18 = 2/3$. Since $T \propto P/n$, the new temperature is $T_2 = T_1 \times \frac{3/5}{2/3} = 300 \times \frac{9}{10} = 270\text{ K}$.

Answer 1

The Correct Option is C

Concept:
This problem applies the Ideal Gas Law: $PV = nRT$. Since the gas is confined in a rigid cylinder, the volume $V$ is constant. The universal gas constant $R$ is also constant. Therefore, we can establish a proportionality between the varying states: $$\frac{P_1}{n_1 T_1} = \frac{P_2}{n_2 T_2}$$
Step 1: Identify the initial state variables (State 1).
Initial moles, $n_1 = 18\text{ moles}$ Initial pressure, $P_1 = 15\text{ atm}$ Initial temperature, $T_1 = 300\text{ K}$
Step 2: Identify the final state variables (State 2).
Final pressure, $P_2 = 9\text{ atm}$ Final moles: 6 moles are withdrawn, so the remaining amount is: $n_2 = 18 - 6 = 12\text{ moles}$ Final temperature, $T_2 = ?$
Step 3: Apply the gas law proportion and solve for $T_2$.
Substitute the values into the proportionality equation: $$\frac{15}{(18)(300)} = \frac{9}{(12)T_2}$$ Simplify the fractions: $$\frac{15}{5400} = \frac{9}{12 T_2}$$ $$\frac{1}{360} = \frac{3}{4 T_2}$$ Cross-multiply to solve for $T_2$: $$4 T_2 = 360 \times 3$$ $$4 T_2 = 1080$$ $$T_2 = \frac{1080}{4} = 270\text{ K}$$