A cylinder of volume 0.280 m 3 contains 12.0 mol of neon gas at 23.4°C. Assume n

Question

A cylinder of volume 0.280 m3 contains 12.0 mol of neon gas at 23.4°C. Assume neon behaves as an ideal gas.

(a) What is the pressure of the gas? (Pa)

(b) Find the internal energy of the gas. (J)

(c) Suppose the gas expands at constant pressure to a volume of 1.000 m3. How much work is done on the gas? (J)

(d) What is the temperature of the gas at the new volume? (K)

(e) Find the internal energy of the gas when its volume is 1.000 m3. (J)

(f) Compute the change in the internal energy during the expansion. (J)

(g) Compute U W. (J)

(h) Must thermal energy be transferred to the gas during the constant pressure expansion or be taken away?

(i) Compute Q, the thermal energy transfer. (J)

(j) What symbolic relationship between Q, U, and W is suggested by the values obtained?

Explanation / Answer

Use ideal gas law:
PV = nRT
=>
P = nRT/V
= 12.0 mol 8.3145 Pam³K¹mol¹ (23.4 + 273) K / 0.280 m³
= 105611 Pa

(b)
Internal energy for an ideal gas is given by:
U = nCvT
The molar heat capacity for a monatomic ideal gas like neon is:
Cv = (3/2)R
Hence,
U = (3/2)nRT
= (3/2) 12.0 mol 8.3145 JK¹mol¹ (23.4 + 273) K
= 44359.52 J

(c)
Assuming reversible operation the work done on the gas is given by the integral
W = - P dV from initial to final volume
For a constant pressure process this simply fies to:
W = - P dV = - PV
=>
W = - 105611Pa (1.00 m³ - 0.28 m³ )
= - 76039.9J

(d)
When you separate variable and constant terms in ideal gas law equation you find:
V/T = nR/P = constant
=>
V_final/T_final = V_initial/T_initial
=>
T_final = T_initial (V_final/V_initial)
= (23.4 + 273)K (1.00 m³ / 0.280 m³)
=1058.57k

(e)
U = (3/2)nRT
= (3/2) 12.0 mol 8.3145 JK¹mol¹ 1058.57
= 158426.64j

(f)
U = U_final - U_initial
= 158426.64j- 44359.52 J
= 114067.12J

(g)
I have no idea what DU 2 W means.

(h)
The internal energy of the gas increases although it does work, i.e. it transfer energy to its surroundings. So there must be an additional energy transfer to the gas, such you get an net positive change in internal energy.

(i)
The heat transferred in a constant pressure process equals change in enthalpy:
Q = H
with the formula for enthalpy of an ideal gas
H = nCpT
and the molar heat capacity for a monatomic ideal gas
Cp = (5/2)R
you get
Q = (5/2)nRT
= (5/2) 12.0 mol 8.3145 JK¹mol¹ (1058.57 K + 293 K )
= 337128.86J

(j)
U = Q + W

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.