Two moles of an ideal gas is heated at constant pressure of one atomsphere from $27^{\circ} C$ to $127^{\circ} C .$ if $C_{v, m} = 20 + 10^{- 2} T$ $J K^{- 1} m o l^{- 1}$ then Heat (q) and Internal energy ( $△ U$ ) for the process are respectively:

6362.8 J, 4700 J

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3037.2 J, 4700 J

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7062.8 J, 5400 J

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3181.4 J, 2350 J

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Solution

The correct option is A 6362.8 J, 4700 J
Calculating $w$ using :
$w = - P △ V = - n R △ T$ (for an ideal gas)
$w = - n R △ T = - 2 \times 8.314 \times 100 = - 1662.8 J$

now molar heat capacity at constant volume is given as
$C_{v} = \frac{d U}{n d T}$ or

$d U = n C_{v} d T$
given, $C_{v} = 20 + 10^{- 2} T$
$T_{1} = 300 K$ and $T_{2} = 400 K$ so,
integrating both sides from $T_{1}$ to $T_{2}$ :

$△ U = n \int_{300}^{400} (20 + 10^{- 2} T) d T = 2 [20 \times T + \frac{10^{- 2}}{2} (T^{2})] = 2 [20 \times (T_{2} - T_{1}) + \frac{10^{- 2}}{2} (T_{2}^{2} - T_{1}^{2})] = 2 [20 \times 100 + \frac{10^{- 2}}{2} (400^{2} - 300^{2})] S o, △ U = 4700 J$

also $△ U = q + w$
$4700 = q - 1662.8 ∴ q = 6362.8 J$

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