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In a process, temperature and volume of one mole of an ideal monoatomic gas are varied according to the relation VT=K, where K is a constant . In this process the temperature of the gas is increased by $\Delta T$ . The amount of heat absorbed by gas is (R is gas constant ) :
Option 1)
$\frac{2K}{3}\Delta T$
Option 2)
$\frac{1}{2} R \Delta T$
Option 3)
$\frac{1}{2} KR \Delta T$
Option 4)
$\frac{3}{2} R \Delta T$

Answers (1)

best_answer

Specific Heat in Polytropic Process -

$C= C_{V}-\frac{R}{r-1}$

$\left ( PV^{r} = constant\right )$

- wherein

$\because C_{V}= \frac{R}{\gamma -1}$

$\because C_{V}= \frac{R}{\gamma -1}- \frac{R}{r -1}$

$\gamma =$ Adiabatic exponent

Given $VT=K\cdots \cdots (1)$

We Know $PV=nRT\cdots \cdots (2)$

From (1) & (2)

$V\cdot \left ( \frac{PV}{nR} \right )=K$

$\Rightarrow PV^{2}=K\cdots (3)$

This is polytropic process with n=2

So far polytropic process

$C=\frac{R}{1-n}+C_{v}=\frac{R}{1-2}+\frac{3R}{2}=\frac{R}{2}$

We know

$\Delta Q=nC\Delta T$

$=n\left ( \frac{R}{2} \right )\Delta T$

$\Delta Q=\frac{nR\Delta T}{2}$

Option 1)

$\frac{2K}{3}\Delta T$

Option 2)
$\frac{1}{2} R \Delta T$
Option 3)

$\frac{1}{2} KR \Delta T$

Option 4)

$\frac{3}{2} R \Delta T$

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