Question

2-A 2.00-mol sample of helium gas initially at 300 K and 0.400 atm is compressed isothermally to 1.20 atm. Assuming the behavior of helium to be that of an ideal gas, find the energy transferred by heat ? a) -5.38 kJ b)-5.28 kJ c) -5.58 kJ d) -5.48 kJ​

Answer 1

Answer:

- 5.48 kg

Explanation:

The final volume will be one-third of the original, for the temperature to be constant, the sample must liberate as many joules by heat as it takes in by work.

The ideal gas law can tell us the original and final volumes. The negative integral of pdV will tell us the work input, and the first law of thermodynamics will tell us the energy output by heat.

1. Rearranging PV=nRT, we get Vi=PinRT

The initial volume is

Vi=(0.400atm)(1.013×105Pa/atm)(2.00mol)(8.314J/molK)(300K)(N/m21Pa)=0.123m3

For isothermal compression, PV is constant, so PiVi=PfVf and the final volume is

Vf=ViPfPi=(0.123m3)(1.20atm0.400atm)=0.0410m3.

2. W=−∫pdV=−∫VnRTdV=−nRTln(ViV