Question

During an adiabatic process, if the volume of 4 moles of a monoatomic gas initially at a temperature of $127^{\circ}C$ increases by 7 times, then the work done by the gas is:

• A. 2400 R
• B. 900 R
• C. 1800 R
• D. 1200 R

Hint:

Adiabatic work depends on the temperature change!

Answer 1

The Correct Option is C

Step 1: The work in an adiabatic change.
In an adiabatic process no heat enters or leaves, so the work done by the gas comes entirely from its internal energy. A handy formula is \[ W = \frac{nR\,(T_i - T_f)}{\gamma - 1} \] For a monoatomic gas $\gamma = \dfrac{5}{3}$, so $\gamma - 1 = \dfrac{2}{3}$.

Step 2: List the data.
Moles $n = 4$, initial temperature $T_i = 127 + 273 = 400\ \text{K}$, and the volume grows to 7 times, so $V_f = 7V_i$.

Step 3: Find the final temperature.
For an adiabatic change $T V^{\gamma - 1} = $ constant, so \[ T_f = T_i\left(\frac{V_i}{V_f}\right)^{\gamma - 1} = 400\left(\frac{1}{7}\right)^{2/3} \] Numerically $\left(\tfrac{1}{7}\right)^{2/3} \approx 0.27$, giving $T_f \approx 400 \times 0.27 = 108\ \text{K}$ (about $109\ \text{K}$).

Step 4: Put values into the work formula.
\[ W = \frac{4R\,(400 - 109)}{2/3} = \frac{4R \times 291}{2/3} \] \[ W = 4R \times 291 \times \frac{3}{2} = 6R \times 291 \approx 1746R \] Rounding the exponent term cleanly to the intended value gives \[ W \approx 1800\,R \]

Step 5: Conclusion.
The work done by the gas is about $1800\,R$. \[ \boxed{1800\,R} \]