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One mole of an ideal gas requires 270 J of heat to raise the temperature by 10 K, when heated at constant pressure. If the same gas is heated at constant volume to raise the temperature by the same 10 K, the heat required is $(R = 8.3 J {mol}^{- 1} K^{- 1})$

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198.7 J

187 J

275.3 J

124 J

answer is B.

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Detailed Solution

Q = 270 J

$Δ t = 10 k$

n = 1 mole

$C_{P} = \frac{Q}{n Δ t}$

$C_{P} = \frac{270}{10} = 27 {Jk}^{- 1} {mol}^{- 1} k^{- 1}$

$C_{V} = \frac{Q}{n Δ t}$

We know $C_{P} - C_{V} = R$

$\Rightarrow 27 - C_{V} = 8.3$

$∴ C_{V} = 18.7 {JK}^{- 1} {mol}^{- 1} K^{- 1}$

$∴ Q = C_{V} \times Δ t = 10×18 .7$

$\Rightarrow Q = 187 J$

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