Question

Heat leaks into a vessel, containing one mole of an ideal monoatomic gas at a constant rate of $\frac{R}{4} s^{-1}$ where R is universal gas constant. It is observed that the gas expands at a constant rate $\frac{dV}{dt} = \frac{V_0}{400 (second)}$ where $V_0$ is initial volume. The initial temperature is given by $T_0 = 40 K$. What will be final temperature of gas at time t = 400 s.

• A. 40 K
• B. 80 K
• C. 200 K
• D. 260 K

Answer 1

Answer: (B)

80 K

Answer 2

The problem describes a thermodynamic process for one mole of an ideal monoatomic gas. We are given:

1. Amount of gas: $n = 1$ mole.
2. Gas type: Ideal monoatomic.
3. Rate of heat leak into the vessel: $\frac{dQ}{dt} = \frac{R}{4}$ (assuming $R$ is the universal gas constant and the rate is in Watts, i.e., J/s).
4. Rate of volume expansion: $\frac{dV}{dt} = \frac{V_0}{400}$ (assuming $V_0$ is the initial volume and the rate is in $m^3/s$).
5. Initial temperature: $T_0 = 40$ K.
6. Initial volume: $V_0$.
7. Time duration: $t = 400$ s.

We use the First Law of Thermodynamics: $\Delta Q = \Delta U + \Delta W$. Differentiating with respect to time, we get: $\frac{dQ}{dt} = \frac{dU}{dt} + \frac{dW}{dt}$

For an ideal monoatomic gas, the internal energy $U = \frac{3}{2} nRT$. So, the rate of change of internal energy is: $\frac{dU}{dt} = \frac{3}{2} nR \frac{dT}{dt}$ Since $n=1$, $\frac{dU}{dt} = \frac{3}{2} R \frac{dT}{dt}$.

The rate of work done by the gas is $\frac{dW}{dt} = P \frac{dV}{dt}$. Using the ideal gas law, $PV = nRT$, we have $P = \frac{nRT}{V}$. Substituting this into the work rate equation: $\frac{dW}{dt} = \frac{nRT}{V} \frac{dV}{dt}$ Since $n=1$, $\frac{dW}{dt} = \frac{RT}{V} \frac{dV}{dt}$.

Now, substitute these into the First Law equation: $\frac{R}{4} = \frac{3}{2} R \frac{dT}{dt} + \frac{RT}{V} \frac{dV}{dt}$

We can divide the entire equation by $R$: $\frac{1}{4} = \frac{3}{2} \frac{dT}{dt} + \frac{T}{V} \frac{dV}{dt}$

We are given that $\frac{dV}{dt} = \frac{V_0}{400}$, which is a constant rate. Integrating this to find the volume $V(t)$ at any time $t$: $V(t) = V_0 + \frac{V_0}{400} t = V_0 \left(1 + \frac{t}{400}\right)$

Substitute $V(t)$ into the energy equation: $\frac{1}{4} = \frac{3}{2} \frac{dT}{dt} + \frac{T}{V_0 \left(1 + \frac{t}{400}\right)} \left(\frac{V_0}{400}\right)$ $\frac{1}{4} = \frac{3}{2} \frac{dT}{dt} + \frac{T}{\frac{400+t}{400}} \left(\frac{1}{400}\right)$ $\frac{1}{4} = \frac{3}{2} \frac{dT}{dt} + \frac{T}{400+t}$

Rearranging this first-order linear differential equation: $\frac{dT}{dt} + \frac{2}{3(400+t)} T = \frac{1}{6}$

The integrating factor is $(400+t)^{2/3}$. Multiplying by the IF: $\frac{d}{dt} \left[ T (400+t)^{2/3} \right] = \frac{1}{6} (400+t)^{2/3}$

Integrating both sides: $T (400+t)^{2/3} = \frac{1}{10} (400+t)^{5/3} + C$

Using the initial condition $T(0) = 40$ K: $40 (400)^{2/3} = \frac{1}{10} (400)^{5/3} + C \implies C = 0$.

So, $T(t) = \frac{1}{10} (400+t)$. At $t = 400$ s: $T(400) = \frac{1}{10} (400 + 400) = \frac{800}{10} = 80$ K.

This process is isobaric because $V(t) = V_0 \frac{400+t}{400}$ and $T(t) = \frac{400+t}{10}$, which implies $V \propto T$. For an isobaric process, $V_f = 2V_0$ leads to $T_f = 2T_0 = 2 \times 40 \text{ K} = 80 \text{ K}$. The heat input rate for an isobaric process is $\frac{dQ}{dt} = n C_P \frac{dT}{dt} = (1) (\frac{5}{2}R) (\frac{1}{10}) = \frac{R}{4}$, which matches the given rate.