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# Isothermal compression? n = 2.62 mol of Hydrogen gas is initially at T = 336 K temperature and pi = 3.43?

n = 2.62 mol of Hydrogen gas is initially at T = 336 K temperature and pi = 3.43×10^5 Pa pressure. The gas is then reversibly and isothermally compressed until its pressure reaches pf = 9.01×10^5 Pa.

A) What is the volume of the gas at the end of the compression process?

B) How much work did the external force perform?

C) How much heat did the gas emit?

D) How much entropy did the gas emit?

E) What would be the temperature of the gas, if the gas was allowed to adiabatically expand back to its original pressure?

### 1 Answer

- Lee Wang TuiLv 77 years agoFavorite Answer
n = 2.62

T = 336 K

pi = 3.443 bar

pf = 9.01 bar

A)

pV = nRT

pV/nT = R = constant

p1V1/n1T1 = p2V2/n2T2, T1 = T2 and n1 = n2

V2 = p1V1/p2

with, V1 = n1RT1/p1

B)

p1V1/n1T1 = p2V2/n2T2, T1 = T2 and n1 = n2

p1V1 = p2V2 = constant

pv = constant = C

p = C/v

Work,

W = ∫pdv

W = C∫ dv/v

W = C ln v | v1 to v2

W = (p1V1) * ln(V2/V1)

C)

Q = U2-U1 = Q - W

Q = m(u2-u1) + W

Q = m*cv*(T2-T1) + W, T2 = T1, assume contant specific heat

Q = W

D) Tds = Q

S2-S1 = Q/T

E) Adibatic process and reversible, hence the gas undergoes isentropic process.

state 2 and p1 are known.

p1v1/T1 = p2V2/T2

v1/T1 = p2V2/T2p1..... (1)

Hydrogen is diatomic gas, cp/cv = k = 1.4

Isentropic process,

pv^k = constant = c

p1v1^k = p2v2^k ....(2)

Subtitute (2) to (1) to solve for T1

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