Work done in Thermo!

A mono-atomic gas undergoes a process in which it uses 50% of heat in work done by gas. If the process equation of gas is V P k VP^k = constant then value of ' k k ' will be


The answer is 3.

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2 solutions

Tanishq Varshney
May 30, 2015

For a process P V x = c o n s t a n t PV^{x}=constant

molar specific heat capacity C = R γ 1 + R 1 x \large{C=\frac{R}{\gamma-1}+\frac{R}{1-x}}

W = n R d T 1 x W=\frac{nRdT}{1-x}

where γ \gamma is ratio of specific heat capacities

H = n C d T H=nCdT and here x = 1 k x=\frac{1}{k} and γ = 5 3 \gamma=\frac{5}{3}

Given in the question H 2 = W \frac{H}{2}=W

on solving

k k 1 = 3 2 \frac{k}{k-1}=\frac{3}{2}

k = 3 k=3

50 % 50 \text{\%} of heat is used up as work done by gas. 50 % \Rightarrow 50 \text{\%} of heat is used in increasing the internal energy of the gas.

Δ U = 1 2 Δ Q \Rightarrow \Delta U = \frac{1}{2} \Delta Q

Δ U = n C v Δ T = n R γ 1 Δ T \Delta U = n C_v \Delta T = n \frac{R}{\gamma - 1} \Delta T and Δ Q = n C Δ T , C = R γ 1 + R 1 x \Delta Q = n C \Delta T , C = \frac{R}{\gamma - 1} + \frac{R}{1 - x} where x x is polytropic factor.

n R γ 1 Δ T = 1 2 × n ( R γ 1 + R 1 x ) Δ T n \frac{R}{\gamma - 1} \Delta T = \frac{1}{2} \times n( \frac{R}{\gamma - 1} + \frac{R}{1 - x} )\Delta T

R γ 1 = R 1 x \frac{R}{\gamma - 1} = \frac{R}{1-x}

γ 1 = 1 x \gamma -1 = 1 - x

x = 2 γ = 2 5 3 = 1 3 x = 2 - \gamma = 2 - \frac{5}{3} = \frac{1}{3}

The process is P V x = k P V^x = k and x = 1 3 P 3 V = k 3 = c o n s t . x = \frac{1}{3} \Rightarrow P^3 V = k^3 = const.

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