At a temperature of -33.0 degrees C, a sample of confined gas exerts a pressure of 53.3 kPa. If volume remains constant, at what temperature will the pressure reach 133 kPa?

598.9K = 325.9°C

Explanation:

P1= 53.3kPa T1= -33+273=240K

P2=133kPa , T2= ?

Applying P1/T1 = P2/T2

Substitute and Simplify

53.3/240 = 133/T2

T2= 598.9K = 325.9°C

vanderwaal's forces of attraction are the electrostatic intermolecular forces that exist in a molecule due to the temporary or permanent dipoles that are developed in a molecule. in , the major intermolecular forces are dipole-dipole interactions as the molecule is polar (bent structure). so the vanderwaal's forces of interaction in between molecules are the permanent dipole-dipole interactions that develop due to the dipole that exists in between the polar h - s bond. the positively charged hydrogen atom is attracted towards the negative end of the dipole that is sulfur of the adjacent molecule. therefore, these forces of attraction are relatively stronger than that of london dispersion forces that exist in non polar molecules.

your answer is a. the enthalpy of formation for a pure element in its standard state is always positive.