A fixed mass of helium gas is made to undergo a process in which its pressure varies linearly from 1 kPa to 2 kPa, in relation to its volume as the latter varies from 0.2 $m^{3}$ to 0.4 $m^{3}$ . The heat absorbed by the gas will be

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300 J

900 J

1200 J

1500 J

answer is C.

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Detailed Solution

$∆ W = (1 + 2) \times 10^{3} N / m^{2} \times (0.4 - 0.2) m^{3} = 300 J$

$Δ U = n C_{V} ∆ T =$ $\frac{(P_{2} V_{2} - P_{1} V_{1})}{γ - 1}$ $\begin{array}{l} = (3 / 2) (0.4 \times 2 - 0.2 \times 1) \times 103 J = 900 J \\ Δ Q = Δ W + Δ U = 1200 J \end{array}$

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