An ideal gas is taken from state 1 to state 2 through a process represented by a straight line on a Pressure (P) versus Volume (V) diagram. The coordinates (V, P) of state 1 are (10 L , 20 kPa ) and of state 2 are (30 L , 60 kPa ) . If the heat supplied to the gas during this process is 2000 J , what is the change in its internal energy?

Step-by-step solution

The work done by the gas is the area under the P-V graph from state 1 to state 2. Since the process is represented by a straight line, the area under the curve forms a trapezium. Work done, W = 1 2 (P₁ + P₂) (V₂ - V₁) Given P₁ = 20 kPa , P₂ = 60 kPa , V₁ = 10 L , and V₂ = 30 L . W = 1 2 (20 + 60) 10^3 Pa (30 - 10) 10⁻³ m ^3 W = 1 2 80 20 J = 800 J According to the first law of thermodynamics: Q = U + W Given the heat supplied to the system, Q = 2000 J . 2000 J = U + 800 J U = 1200 J Answer: 1200 J