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QPages 77-80: Q19: HC Verma: Concepts of Physics - Volume 2 Specific Heat Capacities of Gases Pages 77-80: Q19 Solution: A gas is enclosed in a cylindrical can fitted with a piston.…

⚠️ Common Exam Traps & Conceptual Pitfalls

Students often apply the reversible adiabatic relation $PV^{\gamma}=\text{constant}$ to a sudden (irreversible) compression, assuming the same $P_2$ and $T_2$ as in a quasistatic process. In reality only a reversible adiabatic obeys $PV^{\gamma}=\text{const}$; for an abrupt compression you must use the first law $\Delta U = -W$ with $W = P_{\text{ext}}\Delta V$ to find the final temperature and pressure.

A gas is enclosed in a cylindrical can fitted with a piston. The walls of the can and the piston are adiabatic. The initial pressure, volume and temperature of the gas are 100 kPa,

400 \text{ cm}^3

and 300 K, respectively. The ratio of the specific heat capacities of the gas,

C_p / C_v = 1.5

. Find the pressure and the temperature of the gas if it is a) suddenly compressed b) slowly compressed to

100 \text{ cm}^3

.

Step-by-Step Verified Solution

**Answer:** (a)

P_2 = 800

kPa,

T_2 = 600

K (b) Same values. **Step-by-Step Explanation:** Given:

P_1 = 100

kPa,

V_1 = 400

cm_,

T_1 = 300

K,

\gamma = 1.5

,

V_2 = 100

cm_. (a) Suddenly compressed (adiabatic):

P_2 = P_1(4)^{1.5} = 800

kPa.

T_2 = T_1(4)^{0.5} = 600

K. (b) Slowly compressed but walls adiabatic, so still adiabatic: same values [citation:16].

All Questions in this Chapter

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