An ideal gas enters a 4cm diameter pipe at a temperature of 35°c and a pressure of 250k pa. after some distance the gas exits the 4cm pipe at 35°c and 105k pa. the specific molar volume (volume per mole) of the inlet gas is the specific molar volume of the outlet gas. (a) greater than (b) less than (c) the same as (d) no enough information to answer

b) volume per mol (10.237 L/mol) of the inlet gas less than of the outlet gas ( 24.378 L/mol ).

Explanation:

∴ R = 0.082 atm*L / K*mol...ideal gas constant

for entry:

T1 = 35°C + 273 = 308 K

P1 = 250 KPa * (  0.009869 atm / KPa) = 2.467 atm

⇒(V/n)1 = (R * T1 ) / P1  = (( 0.082 atm*L / K*mol ) * (308 K )) / 2.467 atm

⇒ (V/n)1 = 10.237 L/mol  ideal gas inlet

for the exit:

T2 = 35 °C + 273 = 308 K

P2 = 105 KPa * ( 0.009869 atm / KPa ) = 1.036 atm

⇒ (V/n)2 = (R * T2) / P2 = (( 0.082 * 308 )) / 1.036

⇒ (V/n)2 = 24.378 L/mol ideal gas outlet

question: ray weighs 100 kilograms, his wrestling partner, bobby, weighs 50 kilograms. if ray moves at 4 m s , how fast will bobby have to run at him to apply an equal amount of kinetic energy?

answer: 8 m/s he needs to run twice as fast to keep up so he has some work to do

explanation:   if the mass has units of kilograms and the velocity of meters per second, the kinetic energy has units of kilograms-meters squared per second squared.

question answered by

(jacemorris04)