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(C) The heat absorbed by the gas is given by the first law of thermodynamics: $Q = \Delta U + W$.For a monoatomic gas, the change in internal energy i

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$16$

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(C) The heat absorbed by the gas is given by the first law of thermodynamics: $Q = \Delta U + W$.For a monoatomic gas, the change in internal energy is $\Delta U = n C_{V} \Delta T = n \left(\frac{3}{2} Right) \Delta T = \frac{3}{2} (P_{f} V_{f} - P_{i} V_{i})$.Given initial state $(P_{i}, V_{i}) = (P_{o}, V_{o})$ and final state $(P_{f}, V_{f}) = (3 P_{o}, 3 V_{o})$.$\Delta U = \frac{3}{2} (3 P_{o} \cdot 3 V_{o} - P_{o} V_{o}) = \frac{3}{2} (9 P_{o} V_{o} - P_{o} V_{o}) = \frac{3}{2} (8 P_{o} V_{o}) = 12 P_{o} V_{o}$.The work done $W$ is the area under the $P-V$ graph. Since the pressure changes linearly with volume from $(P_{o}, V_{o})$ to $(3 P_{o}, 3 V_{o})$, the area is a trapezoid.$W = 	ext{Area} = \frac{1}{2} (P_{i} + P_{f}) (V_{f} - V_{i}) = \frac{1}{2} (P_{o} + 3 P_{o}) (3 V_{o} - V_{o}) = \frac{1}{2} (4 P_{o}) (2 V_{o}) = 4 P_{o} V_{o}$.Thus, the total heat absorbed is $Q = \Delta U + W = 12 P_{o} V_{o} + 4 P_{o} V_{o} = 16 P_{o} V_{o}$.

$A$ quantity of monoatomic gas undergoes a process in which pressure is changed linearly with volume. The pressure and volume are changed from initial value $(P_{o}, V_{o})$ to final value $(3 P_{o}, 3 V_{o})$. The heat absorbed by the gas during the process is (in $P_{o} V_{o}$)

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