SATHEE: Heat and Thermodynamics 5 Question 35

Heat and Thermodynamics 5 Question 35

36. An ideal monoatomic gas is confined in a horizontal cylinder by a spring loaded piston (as shown in the figure). Initially the gas is at temperature $T_{1}$ , pressure $p_{1}$ and volume $V_{1}$ and the spring is in its relaxed state. The gas is then heated very slowly to temperature $T_{2}$ , pressure $p_{2}$ and volume $V_{2}$ . During this process the piston moves out by a distance $x$ . Ignoring the friction between the piston and the cylinder, the correct statements is/are

(2015 Adv.)

(a) If $V_{2} = 2 V_{1}$ and $T_{2} = 3 T_{1}$ ,

then the energy stored in the spring is $\frac{1}{4} p_{1} V_{1}$

(b) If $V_{2} = 2 V_{1}$ and $T_{2} = 3 T_{1}$ ,

then the change in internal energy is $3 p_{1} V_{1}$

then the work done by the gas is $\frac{7}{3} p_{1} V_{1}$

(d) If $V_{2} = 3 V_{1}$ and $T_{2} = 4 T_{1}$ ,

then the heat supplied to the gas is

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Answer:

Correct Answer: 36. $(a, b, c)$

Solution:

Note This question can be solved if right hand side hand side chamber is assumed open, so that its pressure remains constant even if the piston shifts towards right.

(a)

$\begin{matrix} p V = n R T \\ p \propto \frac{T}{V} \end{matrix}$

Temperature is made three times and volume is doubled

$\Rightarrow p_{2} = \frac{3}{2} p_{1}$

Further $x = \frac{Δ V}{A} = \frac{V_{2} - V_{1}}{A} = \frac{2 V_{1} - V_{1}}{A} = \frac{V_{1}}{A}$

$p_{2} = \frac{3 p_{1}}{2} = p_{1} + \frac{k x}{A} \Rightarrow k x = \frac{p_{1} A}{2}$

Energy of spring

$\frac{1}{2} k x^{2} = \frac{p_{1} A}{4} x = \frac{p_{1} V_{1}}{4}$

(b) $Δ U = n c_{v} Δ T = n (\frac{3}{2} R) Δ T$

$\begin{aligned} = \frac{3}{2} (p_{2} V_{2} - p_{1} V_{1}) \\ = \frac{3}{2} [(\frac{3}{2} p_{1}) (2 V_{1}) - p_{1} V_{1}] = 3 p_{1} V_{1} \end{aligned}$

$\Rightarrow p_{2} = \frac{4}{3} p_{1} = p_{1} + \frac{k x}{A}$

$\Rightarrow k x = \frac{p_{1} A}{3}$

$\Rightarrow x = \frac{Δ V}{A} = \frac{2 V_{1}}{A}$

$\begin{aligned} W_{gas} & = (p_{0} Δ V + W_{spring}) = (p_{1} A x + \frac{1}{2} k x \cdot x) \\ = + (p_{1} A \cdot \frac{2 V_{1}}{A} + \frac{1}{2} \cdot \frac{p_{1} A}{3} \cdot \frac{2 V_{1}}{A}) \\ = 2 p_{1} V_{1} + \frac{p_{1} V_{1}}{3} = \frac{7 p_{1} V_{1}}{3} \end{aligned}$

(d) $Δ Q = W + Δ U$

$\begin{aligned} = \frac{7 p_{1} V_{1}}{3} + \frac{3}{2} (p_{2} V_{2} - p_{1} V_{1}) \\ = \frac{7 p_{1} V_{1}}{3} + \frac{3}{2} (\frac{4}{3} p_{1} \cdot 3 V_{1} - p_{1} V_{1}) \\ = \frac{7 p_{1} V_{1}}{3} + \frac{9}{2} p_{1} V_{1} = \frac{41 p_{1} V_{1}}{6} \end{aligned}$

NOTE

$Δ U = \frac{3}{2} (p_{2} V_{2} - p_{1} V_{1})$ has been obtained in part (b).

Heat and Thermodynamics 5 Question 35

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Heat and Thermodynamics 5 Question 35

Answer:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

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