1000cm3(cude) of air at 20C and 101.35kpa is heated at constant pressure until its volume doubles. (A) use the ideal gas equation to calculate the final temprature of the gas? (B) calculate the work done by the gas as it expands.

Answer 1

P V = n R T
P does not change
T1 = 20 + 273 Kelvin
n does not change
R does not change
so
V/T = constant
V2/T2 = V1/T1
2000/T2 = 1000/293
T2 = 2*293 = 586 Kelvin = 313 Centigrade

(B) constant pressure = 101.35 *10^3 Newtons/meter^2
Force on inner surface = 4 pi r^2 * P
work done for change in radius dr = 4 pi P r^2 dr
work done frrom R1 to R2 = 4 pi P (R2^3/3-R1^3/3)
= (4/3) pi P (R2^3/R1^3)
or in other words work done at constant pressure
= P (V2-V1)
V2 = 2000 cm^3 (1 m/10^2 cm)^3 = 2000/10^6 meter^2
so V2 - V1 = 10^3/10^6 = 10^-3 meter^3
so
work done = 101.35*10^3 * 10^-3 = 101.35 Joules

Answer 2

since the has been done by the gas,why could not the answer is negative?

Answer 3

Selam