Question

Work done in compressing adiabatically 1g of air, initially at NTP to one fourth of its original volume is: (take density of air = 0.0001465 g/cm³ and \(\gamma\) = 1.5)

• A. \( 1.365 \times 10^{10} \) ergs
• B. \( 1365 \times 10^8 \) ergs
• C. \( 2.730 \times 10^{10} \) ergs
• D. \( 1.365 \times 10^{10} \) ergs (repeated)

Hint:

In thermodynamics problems, pay close attention to units. The CGS unit of energy is the 'erg' (\(1 \, \text{erg} = 1 \, \text{dyn} \cdot \text{cm}\)). Ensure your pressure is in dyn/cm² and volume is in cm³ to get the work in ergs. Also, be aware of standard approximations for NTP/STP; sometimes 1 atm is taken as \(10^5\) Pa (SI) or \(10^6\) dyn/cm² (CGS) for simplicity.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
The work done on a gas during an adiabatic compression is given by the change in its internal energy. We can calculate this work using the initial and final states (pressure and volume) of the gas. The process is adiabatic, so there is no heat exchange with the surroundings.
Step 2: Key Formula or Approach:
The work done in an adiabatic process from state 1 to state 2 is given by:
\[ W = \frac{P_2 V_2 - P_1 V_1}{\gamma - 1} \] where this formula gives work done *by* the gas. For compression, the work is done *on* the gas, so \(W_{\text{on}} = -W = \frac{P_1 V_1 - P_2 V_2}{\gamma - 1}\).
We also use the adiabatic relation: \( P_1 V_1^\gamma = P_2 V_2^\gamma \).
Step 3: Detailed Explanation:
1. Determine Initial Conditions (State 1):
The gas is initially at NTP (Normal Temperature and Pressure). Mass, \(m = 1\) g. Density, \(\rho = 0.0001465\) g/cm³. Initial Volume, \( V_1 = \frac{m}{\rho} = \frac{1 \, \text{g}}{0.0001465 \, \text{g/cm}^3} \approx 6825.94 \, \text{cm}^3 \).
For NTP, we take \( P_1 = 1 \, \text{atm} \approx 1.013 \times 10^6 \) dyn/cm². To match the answer options, it's common in such problems to approximate \(P_1 = 10^6\) dyn/cm² (1 bar). Let's use this approximation.
\( P_1 = 10^6 \) dyn/cm².
2. Determine Final Conditions (State 2):
The gas is compressed to one-fourth of its original volume.
\( V_2 = \frac{V_1}{4} \).
Using the adiabatic relation to find \(P_2\):
\[ P_2 = P_1 \left(\frac{V_1}{V_2}\right)^\gamma = P_1 (4)^{1.5} = P_1 (4^{3/2}) = P_1 (2^3) = 8P_1 \] \[ P_2 = 8 \times 10^6 \, \text{dyn/cm}^2 \] 3. Calculate the Work Done:
We are calculating the work done *on* the gas (compression).
\[ W_{\text{on}} = \frac{P_2 V_2 - P_1 V_1}{\gamma - 1} \] Substitute \(P_2 = 8P_1\) and \(V_2 = V_1/4\):
\[ W_{\text{on}} = \frac{(8P_1) (\frac{V_1}{4}) - P_1 V_1}{1.5 - 1} = \frac{2P_1 V_1 - P_1 V_1}{0.5} = \frac{P_1 V_1}{0.5} = 2 P_1 V_1 \] Now, plug in the values:
\[ W_{\text{on}} = 2 \times (10^6 \, \text{dyn/cm}^2) \times (6825.94 \, \text{cm}^3) \] \[ W_{\text{on}} = 13651880000 \, \text{ergs} \approx 1.365 \times 10^{10} \, \text{ergs} \] Step 4: Final Answer:
The work done in compressing the air is \( 1.365 \times 10^{10} \) ergs.