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Heat and Work

One mole of a...

Question

One mole of an ideal gas $(C_{v} = 20 J K^{- 1} m o l^{- 1})$ initially at STP is heated at constant volume to twice the initial temperature. For the process, work done ( $w$ ) and $q$ will be :

w = 0; q = 5.46 k J

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w = 0; q = 0

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w = - 5.46 k J; q = 5.46 k J

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w = 5.46 k J; q = 5.46 k J

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Solution

The correct option is A $w = 0; q = 5.46 k J$
As volume is constant so work done is zero,
$w = - P △ V = P \times 0 = 0$
heat absorbed at constant volume is given as
$q_{v} = n C_{v} (△ T)$
$n = 1$
$C_{v} = 20 J K^{- 1} m o l^{- 1}$
now $△ T = T_{2} - T_{1} = 546 - 273 = 273 K$
(as at STP, $T = 273 K$ )
so,
$q_{v} = 20 \times 273 = 5460 J = 5.46 k J$

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Heat and Work

Standard XII Chemistry

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One mole of an ideal gas (Cv=20 JK−1mol−1) initially at STP is heated at constant volume to twice the initial temperature. For the process, work done (w) and q will be :

The correct option is A w=0; q=5.46 kJAs volume is constant so work done is zero, w=−P△V=P×0=0 heat absorbed at constant volume is given as qv=nCv(△T) n=1 Cv=20 JK−1mol−1 now △T=T2−T1=546−273=273 K (as at STP, T=273 K) so, qv=20×273=5460 J=5.46 kJ

One mole of an ideal gas $(C_{v} = 20 J K^{- 1} m o l^{- 1})$ initially at STP is heated at constant volume to twice the initial temperature. For the process, work done ( $w$ ) and $q$ will be :