Question

A cylindrical vessel contains 16 kg of gas at a pressure of 1 atmosphere. A certain amount of gas is taken out and the pressure of gas in the vessel becomes 0.75 atmosphere. The amount of gas taken out is

• A. \(2.5\,\text{kg} \)
• B. \(4\,\text{kg} \)
• C. \(7.5\,\text{kg} \)
• D. \(8.25\,\text{kg} \)
• E. \(10\,\text{kg} \)

Hint:

At constant \(T,V\), pressure directly proportional to amount of gas.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
This problem can be solved using the Ideal Gas Law, \(PV=nRT\). Since the gas is in a cylindrical vessel, its volume (V) is constant. We can also assume that the process of taking gas out happens at a constant temperature (T). Under these conditions, the pressure (P) of the gas is directly proportional to the number of moles (n), which is in turn directly proportional to the mass (m) of the gas.
Step 2: Key Formula or Approach:
From the Ideal Gas Law, with V and T constant, we have \(P \propto n\). Since the number of moles \(n = \frac{\text{mass (m)}}{\text{Molar Mass (M)}}\), and M is constant for a given gas, we have \(n \propto m\). Therefore, \(P \propto m\). This implies that the ratio of pressures is equal to the ratio of masses: \[ \frac{P_1}{P_2} = \frac{m_1}{m_2} \] Step 3: Detailed Explanation:
Let the initial state be state 1 and the final state be state 2. We are given:
Initial mass, \(m_1 = 16\) kg.
Initial pressure, \(P_1 = 1\) atm.
Final pressure, \(P_2 = 0.75\) atm.
We need to find the final mass, \(m_2\), remaining in the vessel. Using the proportionality relationship: \[ \frac{P_1}{P_2} = \frac{m_1}{m_2} \] \[ \frac{1}{0.75} = \frac{16}{m_2} \] Solve for \(m_2\): \[ m_2 = 16 \times 0.75 = 16 \times \frac{3}{4} = 4 \times 3 = 12 \text{ kg} \] This is the mass of the gas remaining in the vessel. The question asks for the amount of gas taken out. Amount taken out = Initial mass - Final mass \[ \Delta m = m_1 - m_2 = 16 \text{ kg} - 12 \text{ kg} = 4 \text{ kg} \] Step 4: Final Answer:
The amount of gas taken out is 4 kg.