Question

14.A fixed mass of an ideal gas absorbs 1000J of heat and expands under a constant pressure of 20 kPa from a volume of 25 x10-3 m3 to a volume of 50 x10-3 m³ What is the change in internal energy of the gas?

Answer 1

The heat absorbed by the ideal gas [dQ] = 1000 J

The initial volume of the ideal gas $V_{1} =25*10^{-3}\ m^3$

The final volume of the ideal gas   $V_{2} =50*10^{-3}\ m^3$

Hence, the change in volume $dV=V_{2} -v_{1}$

                                                          $=[50*10^{-3} -25*10^{-3}]\ m^3$

                                                          $=25*10^{-3}\ m^3$

The pressure applied on the gas $P=20\ kPa$

                                                   $i.e\ P=20*10^3\ Pa$

The work done by the ideal gas dW = PdV

                                                            =$20*10^3*25*10^{-3}\ J$

                                                            = 500 J

We are asked to calculate the change in internal energy of the gas.

From first law of thermodynamics, we know that -

                                       dQ= dU + dW

                                    ⇒dU = dQ-dW

                                             = 1000 J-500 J

                                              = 500 J.

Hence, the change in internal energy is 500 J.