Question

A.) 1 L of oxygen gas has a temperature of 0

Answer 1

A. P= 1 atm,   T= 0^oc   =273K,   V= 1 L = 0.001 m³, k= 1.38x10^-23

The total translational kinetic energy of molecules of oxygen gas

                     E= 3/2 x kT

                           = 1.5 x 1.38x10^-23 x 273

                      E= 5.65x10^-21 J

B. i. P =1 atm =1.01x10⁵ Pa,          V= 10L =0.01 m³,             N= 1 mole

ideal gas relation            PV= NRT

                                    T= PV/NR

                                        =1.01x10⁵ x 0.01 / ( 1 x 8.3)

                                   T= 121.69 K

The temperature of the gas T= 121.69 K

ii. V₁ = 10 L       V₂= 20L , T^₁= 121.69K    T₂=?   Gama= 1.4 for diatomic gas

since it is Isobaric Process

                          T_1/T₂= (V₂/V₁)^gama-1

                               T₂ = T₁ (V₁/V₂)^{gama -1}

                                     = 121.69 x (10/20)^1.4-1

                                       =121.69 x 0.5^0.4

                                 T₂= 93K

iii.   V= 20L,   T₁= 93K    T₂= 350K, P₁= 1atm , P₂= ?

     It is Isochoric process

                   T₁/T₂ =P1/P₂) ^{(gama-1)/gama}

                       P₂ = P₁ x( T₂/T₁)^1.4/0.4

                              = 1 x (350/ 93)^3.5

                          P₂= 103.4 atm