Question

When 80 J of heat is absorbed by a monotonic gas, its volume increases by \( 16 \times 10^5 \) \( m^3 \). The pressure of the gas is:

• A. \( 2 \times 10^5 \) \( Nm^{-2} \)
• B. \( 4 \times 10^5 \) \( Nm^{-2} \)
• C. \( 6 \times 10^5 \) \( Nm^{-2} \)
• D. \( 5 \times 10^5 \) \( Nm^{-2} \)

Hint:

In thermodynamic processes, work done by a gas can be calculated using \( W = P \Delta V \), and the first law of thermodynamics is crucial for solving heat and work-related problems.

Answer 1

The Correct Option is A

Step 1: Using the First Law of Thermodynamics
The first law of thermodynamics states: \[ \Delta Q = \Delta U + W \] where: - \( \Delta Q = 80 \) J (heat absorbed), - \( \Delta U \) is the internal energy change, - \( W \) is the work done by the gas. For a monotonic ideal gas, the internal energy change is given by: \[ \Delta U = \frac{3}{2} n R \Delta T \] But, since work done by a gas in an isothermal expansion is: \[ W = P \Delta V \] Step 2: Finding the Pressure
From the first law: \[ 80 = P \times (16 \times 10^5) \] Solving for \( P \): \[ P = \frac{80}{16 \times 10^5} \] \[ P = \frac{80}{16} \times 10^{-5} \] \[ P = 5 \times 10^{-5} \times 10^5 \] \[ P = 2 \times 10^5 \text{ Nm}^{-2} \]