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QPages 77-80: Q17: HC Verma: Concepts of Physics - Volume 2 Specific Heat Capacities of Gases Pages 77-80: Q17 Solution: An ideal gas at pressure and temperature 300 K occupies 100 cc. …

⚠️ Common Exam Traps & Conceptual Pitfalls

In an adiabatic compression the gas does negative work. The integral $W=int P,dV$ gives a negative value because $dV<0$. The correct formula is $W=(P_2V_2-P_1V_1)/(gamma-1)$, which yields a negative number; the magnitude is $|W|$ if only the amount of work is required.

An ideal gas at pressure 2.5×105 Pa2.5 \times 10^5 \text{ Pa} and temperature 300 K occupies 100 cc. It is adiabatically compressed to half its original volume. Calculate a) the final pressure b) the final temperature and c) the work done by the gas in the process. Take γ=1.5\gamma = 1.5.

Step-by-Step Verified Solution

**Answer:** (a) P2=7.1×105P_2 = 7.1 \times 10^5 Pa (b) T2=424T_2 = 424 K (c) W=21W = 21 J. **Step-by-Step Explanation:** Given: P1=2.5×105P_1 = 2.5 \times 10^5 Pa, V1=100V_1 = 100 cm_, T1=300T_1 = 300 K, γ=1.5\gamma = 1.5, V2=50V_2 = 50 cm_. (a) P2=P1(V1V2)γ=2.5×105×21.5=7.07×105P_2 = P_1\left(\frac{V_1}{V_2}\right)^\gamma = 2.5 \times 10^5 \times 2^{1.5} = 7.07 \times 10^5 Pa. (b) T2=T1(V1V2)γ−1=300×20.5=424T_2 = T_1\left(\frac{V_1}{V_2}\right)^{\gamma-1} = 300 \times 2^{0.5} = 424 K. (c) W=P1V1−P2V2γ−1=2.5×105×100×10−6−7.07×105×50×10−60.5=21W = \frac{P_1V_1 - P_2V_2}{\gamma-1} = \frac{2.5 \times 10^5 \times 100 \times 10^{-6} - 7.07 \times 10^5 \times 50 \times 10^{-6}}{0.5} = 21 J.

All Questions in this Chapter

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