3 Marks Question

Question 13 Marks

Consider a P-V diagram in which the path followed by one mole of perfect gas in a cylindrical container is shown in Fig.

Given the internal energy for one mole of gas at temperature T is (3/2) RT, find the heat supplied to the gas when it is taken from state (1) to (2) with V₂ = 2V_1.

Answer

$\therefore\ \text{PV}=$ constant = K (given) or $\text{P}_1\text{V}_1^\frac{1}{2}=\text{P}_2\text{V}_2^\frac{1}{2}= \text{K}\ \text{and}\ \text{P}=\frac{\text{K}}{\text{V}^\frac{1}{2}}$

Given that internal energy U of gas is

$\text{U}=\Big(\frac{3}{2}\Big)\text{RT}$

$\Delta\text{U}=\frac{3}{2}\text{RdT}=\frac{3}{2}\text{R}(\text{T}_2-\text{T}_1)$

$\therefore\ \text{T}_2=\sqrt2\text{T}_1,$ from part (b)

$\Delta\text{U}=\frac{3}{2}\text{R}\big[\sqrt2\text{T}_1-\text{T}_1\big]=\frac{3}{2}\text{R}\text{T}_1\big(\sqrt2-1\big)$

Form part (a) $\text{dW}=2\text{P}_1\text{V}^\frac{1}{2}\big(\sqrt{\text{V}_2}-\sqrt{\text{V}_1}\big)$

$\because\ \text{V}_2=2\text{V}_1$ (given)

so, $\sqrt{\text{V}_2}=\sqrt2\sqrt{\text{V}_1}$$$ then

$\text{dW}=2\text{P}_1\text{V}_1^\frac{1}{2}\big(\sqrt2\sqrt{\text{V}_1}-\sqrt{\text{V}_1}\big)$

$=2\text{P}_1\text{V}_1\sqrt{\text{V}_1}\big[\sqrt2-1\big]$

$\text{dW}=2\text{P}_1\text{V}_1\big(\sqrt2-1\big)$

$\text{dW}=2\text{n}\text{R}\text{T}_1\big(\sqrt2-1\big)$ $\big(\therefore\ \text{P}_1\text{V}_1=\text{n}\text{R}\text{T}_1\big)$

$\therefore\ \text{n}=1\therefore\ \text{dW}=2\text{R}\text{T}_1\big(\sqrt2-1\big)$

$\therefore\ \text{dQ}=\text{dW}+\text{dU}=2\text{R}\text{T}_1\big(\sqrt2-1\big)+\frac{3}{2}\text{R}\text{T}_1\big(\sqrt2-1\big)$

$=\big(\sqrt2-1\big)\text{R}\text{T}_1\Big[2+\frac{3}{2}\Big]$

$\text{dQ}=-\big(\sqrt2-1\big)\text{RT} .$

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Question 23 Marks

The initial state of a certain gas is (P_i , V_i , T_i ). It undergoes expansion till its volume becoms V_f . Consider the following two cases:

The expansion takes place at constant temperature.
The expansion takes place at constant pressure.

Plot the P-V diagram for each case. In which of the two cases, is the work done by the gas more?

Answer

The expension from V_i to Vf tempreature T_i remains constant so isothermal expension i.e. P_iV_i = P_fV_f constant T.
The expension is at constant pressure p_i so isobaric process so graph P-V will be parallel to V axis till its volume becomes V_f As the area enclosed by graph (a) is less than (b) with volume axis so W.D. by process (b) is more than of (a).

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Question 33 Marks

Is it possible to increase the temperature of a gas without adding heat to it? Explain.

Answer

Yes, it is possible to increase the temperature of a gas without adding heat to it, during adiabatic compression the temperature of a gas increases while no heat is given to it. For an adiabatic compression, no heat is given or taken out in adiabatic process.

Therefore,$\Delta\text{Q}=0$

According to the first law of thermodynamics,

$\Delta\text{Q}=\Delta\text{U}+\Delta\text{W}$

$\Delta\text{U}=-\Delta\text{W}(\Delta\text{Q}=0)$

In compression work is done on the gas, i.e. work done is negativ. Therefore, $\Delta\text{U}=$ positive.

Hence, internal energy of the gas increases due to which its temperature increases.

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Question 43 Marks

Consider a Carnot’s cycle operating between T₁ = 500K and T₂=300K producing 1k J of mechanical work per cycle. Find the heat transferred to the engine by the reservoirs.

Answer

Efficiency of Carnot's engine $\mu=1-\frac{\text{T}_{2}}{\text{T}_{1}}$

Tempreature of source or reservior = T₁ = 500K

Tempreature of sink T₂ = 300K

$\therefore\mu=1-\frac{\text{T}_{2}}{\text{T}_{1}}$

$\frac{\text{Output work}}{\text{Input work}(E)}=1-\frac{300}{500}$

$\frac{1000\text{J}}{\text{x}}=1-0.6$

$\text{x}=\frac{1000}{\text{x}}=0.4$

$\text{x}=\frac{1000}{0.4}=2500 \text{J} .$

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