Question

An ideal monoatomic gas is confined in a cylinder by a spring loaded piston of cros section $8.0 \times 10^{-3} \, m^2$. Initially the gas is at 300K and occupies a volume of $2.4 \times 10^{-3} \, m^3$ and the spring is in its relaxed state as shown in figure. The gas is heated by a small heater until the piston moves out slowly by $0.1\, m$. The force constant of the spring is $8000\, N/m$ and the atmospheric pressure is $1.0 \times 10^5 \, N/m^2$. The cylinder and the piston are thermally insulated. The piston and the spring are massless and there is no friction between the piston and the cylinder. The final temperature of the gas will be : (Neglect the heat loss through the lead wires of the heater. The heat capacity of the heater coil is also negligible)

• A. 300 K
• B. 800 K
• C. 500 K
• D. 1000 K

Answer 1

The Correct Option is B

To determine the final temperature of the gas, we will apply the principles of thermodynamics and mechanics. We know the gas is ideal and monoatomic, and the system is thermally insulated, which indicates an adiabatic process.

The initial setup:

• Piston cross-sectional area, A = 8.0 \times 10^{-3} \, m^2
• Initial volume, V_1 = 2.4 \times 10^{-3} \, m^3
• Initial temperature, T_1 = 300 \, K
• Atmospheric pressure, P_0 = 1.0 \times 10^5 \, N/m^2
• Force constant of the spring, k = 8000 \, N/m
• Piston displacement, \Delta x = 0.1\, m

The spring is initially relaxed, so the initial force by the spring is zero. As the gas is heated, the piston moves up by \Delta x = 0.1\, m, compressing the spring and exerting an additional force.

Final volume, V_2 = V_1 + A \cdot \Delta x = 2.4 \times 10^{-3} + (8.0 \times 10^{-3}) \cdot (0.1) = 3.2 \times 10^{-3} \, m^3

The force exerted by the spring at displacement is:

F_{\text{spring}} = k \cdot \Delta x = 8000 \times 0.1 = 800 \, N

The total pressure exerted by the gas in the final state is:

P_{\text{final}} = P_0 + \frac{F_{\text{spring}}}{A} = 1.0 \times 10^5 + \frac{800}{8.0 \times 10^{-3}} = 2.0 \times 10^5 \, N/m^2

Using the adiabatic relation for an ideal monoatomic gas:

P_1 \cdot V_1^\gamma = P_2 \cdot V_2^\gamma, \gamma = \frac{5}{3} for a monoatomic ideal gas.

Substituting the values:

(1.0 \times 10^5) \cdot (2.4 \times 10^{-3})^{5/3} = (2.0 \times 10^5) \cdot (3.2 \times 10^{-3})^{5/3}

Solving the equation leads to:

\frac{T_2}{T_1} = \frac{P_2 \, V_2}{P_1 \, V_1}

Thus,

T_2 = T_1 \cdot \frac{(P_2 \cdot V_2)}{P_1 \cdot V_1} = 300 \cdot \frac{(2.0 \times 10^5) \cdot (3.2 \times 10^{-3})}{(1.0 \times 10^5) \cdot (2.4 \times 10^{-3})} = 800 \, K

Therefore, the final temperature of the gas is 800 K.