An ideal gas expands against a constant external pressure of 2 atmosphere from 20L to 40L and absorbs 10kJ of heat from surrounding. What is the change in the internal energy of the system ?

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Solution

Pressure = 2 ATM
= 2 x 1.013 x10^5 Pa
dQ=10 kJ =10^4 J
dV= 40 L - 20 L=20 L
= 20 x 10^-3 m^3=2x10^-2 m^-3

From first law of thermodynamics
dQ = dU+dW
dU = dQ-dw
=dQ -PdV
=10^4–2x1.013x10^5 x2x10^—2
=10000–4052
=5948J

Hope you understand!

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An ideal gas expands against a constant external pressure of 2 atmosphere from 20L to 40L and absorbs 10kJ of heat from surrounding. What is the change in the internal energy of the system ?

Pressure = 2 ATM = 2 x 1.013 x10^5 Pa dQ=10 kJ =10^4 J dV= 40 L - 20 L=20 L = 20 x 10^-3 m^3=2x10^-2 m^-3 From first law of thermodynamics dQ = dU+dW dU = dQ-dw =dQ -PdV =10^4–2x1.013x10^5 x2x10^—2 =10000–4052 =5948J Hope you understand!

Pressure = 2 ATM
= 2 x 1.013 x10^5 Pa
dQ=10 kJ =10^4 J
dV= 40 L - 20 L=20 L
= 20 x 10^-3 m^3=2x10^-2 m^-3

From first law of thermodynamics
dQ = dU+dW
dU = dQ-dw
=dQ -PdV
=10^4–2x1.013x10^5 x2x10^—2
=10000–4052
=5948J

Hope you understand!