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One mole of an ideal with heat capacity $C_{P}$ at constant pressure undergoes the process $T = T_{0} + α V$ where $T_{0}$ and $α$ are constants. If its volume increases from $V_{1}$ to $V_{2}$ , the amount of heat transferred to the gas is

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$C_{P} R T_{0} l n (\frac{V_{2}}{V_{1}})$

$α C_{P} \frac{(V_{2} - V_{1})}{R T_{0}} l n (\frac{V_{2}}{V_{1}})$

$α C_{P} (V_{2} - V_{1}) + R T_{0} l n (\frac{V_{2}}{V_{1}})$

$R T_{0} l n (\frac{V_{2}}{V_{1}}) - α C_{P} (V_{1} - V_{2})$

answer is C.

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Detailed Solution

Method I: heat transferred to the gas is

$Q = Δ U + W$

$Δ U = (1) C_{V} (T_{2} - T_{1})$

$\Rightarrow$ $Δ U = (1) C_{V} [T_{0} + α V_{2} - T_{0} - α V_{1}]$

$\Rightarrow$ $Δ U = α C_{V} (V_{2} - V_{1}) \dots (1)$

Since, $d W = P d V$

$\Rightarrow$ $d W = \frac{R T}{V} d V$

$\Rightarrow$ $d W = R (\frac{T_{0} + α V}{V}) d V$

$\Rightarrow$ $w = \int d W = R [T_{0} \int_{V_{1}}^{V_{2}} \frac{d V}{V} + α \int_{V_{1}}^{V_{2}} d V]$

$\Rightarrow$ $W = R T_{0} ℓ n (\frac{V_{2}}{V_{1}}) + \propto R (V_{2} - V_{1}) \dots (2)$

So $Q = \propto C_{V} (V_{2} - V_{1}) + R T_{0} ℓ n (\frac{V_{2}}{V_{1}}) + α R (V_{2} - V_{1})$

$\Rightarrow$ $Q = \propto (V_{2} - V_{1}) (C_{V} + R) + R T_{0} ℓ n (\frac{V_{2}}{V_{1}})$

$\Rightarrow$ $Q = R T_{0} ℓ n (\frac{V_{2}}{V_{1}}) + α C_{P} (V_{2} - V_{1})$

Method II: Since $T = T_{0} + α V$

(a) $α = 0,$ then

$T = T_{0} =$ constant

$\Rightarrow$ process is isothermal

Since change in internal energy equals zero for an isothermal process, so

$Q_{1} = (1) R T_{0} \ln (\frac{V_{2}}{V_{1}})$

$\Rightarrow Q_{1} = R T_{0} \ln (\frac{V_{2}}{V_{1}})$

(b) Think $T_{0} = 0,$ then

$T = α V$

where, $α$ is a constant

$\Rightarrow$ Process is isobaric

( $∵$ temperature is directly proportional to volume)

$\Rightarrow$ $Q_{2} = (1) C_{p} (T_{2} - T_{1})$

Since, $T_{2} = T_{0} + α V_{2},$ and $T_{1} = T_{0} + α V_{1}$

$\Rightarrow$ $Q_{2} = α C_{p} (V_{2} - V_{1})$

SUPERIMPOSING THE (a) AND (b), WE GET

$Q = R T_{0} \ln (\frac{V_{2}}{V_{1}}) + α C_{p} (V_{2} - V_{1})$

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