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

Phase 1: Conceptual Foundation:
Work performed on a gas during adiabatic compression equates to its internal energy change. This work can be determined using the gas's initial and final pressure and volume states. As the process is adiabatic, no heat is exchanged with the environment.
Phase 2: Governing Equation and Strategy:
The work done by a gas in an adiabatic process from state 1 to state 2 is calculated using:
\[ W = \frac{P_2 V_2 - P_1 V_1}{\gamma - 1} \]This equation quantifies work done *by* the gas. For compression, work is done *on* the gas, hence \(W_{\text{on}} = -W = \frac{P_1 V_1 - P_2 V_2}{\gamma - 1}\).
The adiabatic relationship \( P_1 V_1^\gamma = P_2 V_2^\gamma \) is also utilized.
Phase 3: Detailed Calculation:
1. Initial State (State 1):
The gas begins 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, \( P_1 = 1 \, \text{atm} \approx 1.013 \times 10^6 \) dyn/cm². To align with typical problem resolutions, we approximate \(P_1 = 10^6\) dyn/cm² (1 bar).\( P_1 = 10^6 \) dyn/cm².
2. Final State (State 2):
The gas volume is reduced to one-fourth of its initial volume.
\( V_2 = \frac{V_1}{4} \).
Determining \(P_2\) via the adiabatic relation:
\[ P_2 = P_1 \left(\frac{V_1}{V_2}\right)^\gamma = P_1 (4)^{1.5} = P_1 (4^{3/2}) = P_1 (2^3) = 8P_1 \]\[ P_2 = 8 \times 10^6 \, \text{dyn/cm}^2 \]3. Work Calculation:
Calculating work performed *on* the gas (compression).\[ W_{\text{on}} = \frac{P_2 V_2 - P_1 V_1}{\gamma - 1} \]Substituting \(P_2 = 8P_1\) and \(V_2 = V_1/4\):
\[ W_{\text{on}} = \frac{(8P_1) (\frac{V_1}{4}) - P_1 V_1}{1.5 - 1} = \frac{2P_1 V_1 - P_1 V_1}{0.5} = \frac{P_1 V_1}{0.5} = 2 P_1 V_1 \]Inserting values:
\[ W_{\text{on}} = 2 \times (10^6 \, \text{dyn/cm}^2) \times (6825.94 \, \text{cm}^3) \]\[ W_{\text{on}} = 13651880000 \, \text{ergs} \approx 1.365 \times 10^{10} \, \text{ergs} \]Phase 4: Conclusion:
The work required for the air compression is \( 1.365 \times 10^{10} \) ergs.