Question

A certain mass of gan as NTP its expendedto3 time volume under adatatic conditions calculated the resultant temperature pressure. R=1.4

Answer 1

188.90  K WILLL be the resultant temperature .

Explanation:

for adiabatic expansion

TVⁿ⁻¹ = CONSTANT  .............eq 1

where n is  specific heat ratio, sometimes denoted by gaama

here n= 1.4  (GIVEN )

now

T 1 = 20 °c   ( as ntp is given )

   = 293.15 K

V2/V1 = 3

using eq 1  for condition 1 and 2

T2/T1 = (V1/V2)ⁿ⁻¹

       = (1/3)ⁿ⁻¹       n = 1.4  GIVEN

      = (1/3)^0.4

      = 0.644

T2 = 0.644 * 293.15

     = 188.904  K