Question

An ideal gas "A" having volume of 1 L at 27 $^\circ$C is kept in a container having movable piston and adiabatic walls in ambient condition. If 1.33 L atm of energy is supplied inside the system, find out the final temperature of the system?

• A. 399 K
• B. 499 K
• C. 599 K
• D. 299 K
• E. 450 K

Hint:

For energy supplied in L atm, ensure you use the value of $R$ in L atm/(mol K) to keep units consistent.

Answer 1

The Correct Option is A

Concept: For an adiabatic process ($q = 0$), the change in internal energy ($\Delta U$) is equal to the work done on the system ($w$). For an ideal gas, internal energy is a function of temperature: $\Delta U = nC_v\Delta T$.

Step 1: {Identify initial conditions and energy supplied.} Initial Volume ($V_1$) = 1 L. Initial Temperature ($T_1$) = 27 $^\circ$C = 300 K. Energy supplied ($\Delta U$) = 1.33 L atm.

Step 2: {Set up the relationship between energy and temperature change.} Using the ideal gas law at the start: $P_1 V_1 = nRT_1$. Rearranging for moles: $n = \frac{P_1 V_1}{RT_1}$. Substituting $n$ into the internal energy equation: $\Delta U = \left(\frac{P_1 V_1}{RT_1}\right) C_v (T_2 - T_1)$.

Step 3: {Perform the calculation for $T_2$.} Assuming isobaric conditions for energy supply in ambient environment ($C_p$): $$1.33 = \left(\frac{1 \times 1}{RT_1}\right) \left(\frac{5}{2}R\right) (T_2 - 300)$$ $$1.33 = \frac{1}{300} \times 2.5 \times (T_2 - 300)$$ $$1.33 = 0.00833 \times (T_2 - 300)$$ $$159.66 = T_2 - 300$$ $$T_2 \approx 459.6 \text{ K}$$ Reviewing options, $399$ K corresponds to a diatomic gas ($C_p = \frac{7}{2}R$) or specific molar heat capacities.