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Specific Heat Capacity

A quantity of...

Question

A quantity of an ideal monoatomic gas consists of $n$ moles initially at temperature $T_{1}$ . The pressure and volume both are then slowly doubled in such a manner that the process traces out a straight line on a $P - V$ diagram. The ratio $\frac{Q}{n R T_{1}}$ for the given thermodynamic process is equal to:
(Q is the heat supplied to the gas)

1.5

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6

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4.5

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3

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Solution

The correct option is B $6$
The pressure and volume of the gas is doubled slowly it means it is a reversible process.
$P_{2} = 2 P_{1}$
$V_{2} = 2 V_{2}$

$\Rightarrow P \propto V \Rightarrow P V^{- 1} = C o n s t a n t$
$x = - 1$

$C = C_{V} + \frac{R}{1 - x} = \frac{3 R}{2} + \frac{R}{2} = 2 R$

Heat supplied is $Δ Q = n C Δ T$

Using ideal gas equation at point $1$ :
$P_{1} V_{1} = n R T_{1}$
At the final state $2$ , (number of moles of gas will remain same)
$P_{2} V_{2} = n R T_{2}$
$(2 P_{1}) (2 V_{1}) = n R T_{2}$
$4 P_{1} V_{1} = n R T_{2}$
$4 (n R T_{1}) = n R T_{2}$
$∴ T_{2} = 4 T_{1}$
$Δ Q = n C Δ T = n (2 R) (T_{2} - T_{1})$
$Q = 6 n R T_{1}$

Therefore, $\frac{Q}{n R T_{1}} = 6$

$\begin{matrix} Why this question ? Tip: In this problem the straight line plot of process on P - V diagram indicates that workdone by gas can be easily obtained by finding area under curve and then it can be used to obtain heat supplied (Q) by applying first law of thermodynomics. \end{matrix}$

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