Question

For an ideal gas in a reversible process (\(\Delta Q = 0\)), volume becomes \(8\) times and temperature becomes \(\dfrac{1}{4}\) times the initial value. Identify the gas:

• A. \( \mathrm{CO_2} \)
• B. \( \mathrm{O_2} \)
• C. \( \mathrm{NH_3} \)
• D. \( \mathrm{He} \)

Hint:

Remember common values of \(\gamma\):
Monoatomic gas: \( \gamma=\frac{5}{3} \)
Diatomic gas: \( \gamma=\frac{7}{5} \)
Polyatomic gas: \( \gamma\approx \frac{4}{3} \)

Answer 1

The Correct Option is D

Concept:
For a reversible adiabatic process (\(\Delta Q=0\)) of an ideal gas: \[ TV^{\gamma-1}=\text{constant} \] where \(\gamma=\dfrac{C_p}{C_v}\).
Step 1: Apply the adiabatic relation. \[ T_1 V_1^{\gamma-1} = T_2 V_2^{\gamma-1} \] Given: \[ V_2 = 8V_1,\qquad T_2=\frac{T_1}{4} \] Substitute: \[ T_1 V_1^{\gamma-1} = \frac{T_1}{4}(8V_1)^{\gamma-1} \]
Step 2: Simplify the equation. Cancel \(T_1 V_1^{\gamma-1}\): \[ 1 = \frac{1}{4}\,8^{\gamma-1} \] \[ 8^{\gamma-1} = 4 \] Write in powers of 2: \[ (2^3)^{\gamma-1} = 2^2 \] \[ 3(\gamma-1)=2 \Rightarrow \gamma-1=\frac{2}{3} \Rightarrow \gamma=\frac{5}{3} \]
Step 3: Identify the gas. \[ \gamma=\frac{5}{3} \] corresponds to a monoatomic ideal gas. Among the options: \[ \mathrm{He} \text{ is monoatomic} \] \[ \boxed{\text{The gas is He}} \]