Question

A flexible balloon contains 0.330 mol of an unknown polyatomic gas. Initially the balloon containing the gas has a volume of 7100 cm3 and a temperature of 24.0 ∘C. The gas first expands isobarically until the volume doubles. Then it expands adiabatically until the temperature returns to its initial value. Assume that the gas may be treated as an ideal gas with Cp=33.26J/mol⋅K and γ=4/3.

A. What is the total heat Q supplied to the gas in the process?B. What is the total change in the internal energy ΔU of the gas?C. What is the total work W done by the gas?D. What is the final volume V?

Answer 1

Answer:

Explanation:

The initial temperature is 27+273= 300K.

The volume doubles. V∝T when pressure is constant (isobaric expansion). So the new temperature is 2x300=600K.

There are different ways to do the above step.  E.g. you could say:

P1V1/T1 = P2V2/T

P1V1/300 = P2(2xV1)/T2

V1 cancels and since P1= P2, P1 and P2 cancel

1/300 =  2/T2

T2 = 600K

but this is much more work than using simple proportion.

For the isobaric (constant pressure) expansion, the heat transferred is:

Q = Cp.n.ΔT

= 33.26 x  0.370 x (600 - 300)

= 3692J

= 3690J to 3 significant figures

When the gas expands adiabatically there is no heat transfer (by definition of adiabatc).

So the total heat transfer is 3690J.

Answer 2

Answer:

• Total heat Q supplied = $3259.81J$

• Total change in the internal energy ΔU = $0J$

• The total work W done = $3259.565J$

• Final volume V = $0.1136m^3$

Explanation:

$V_2 = 2V_1\\\\P_1 = \frac{nRT_1}{V_1}\\\\P_1 = \frac{0.33*8.314*297}{7100*10^{-6}}\\\\P_1 = 1.148*10^5 pa$

$T_2 = T_1(\frac{V_2}{V_1}) = 2T_1$

$Q = nCp(T_2 - T_1)\\\\Q = 0.33 (33.26)(297)\\\\Q = 3259.81J$

Workdone,

$w = P(V_2 - V_1)\\\\w = P(2V_1 - V_1)\\\\w = 1.148*10^5(7100*10^{-6})\\\\w = 815.1J$

Internal energy,

$U = Q - w\\\\U = 3259.81 - 815.1\\\\U = 2444.71J$

Change in internal energy,

$U^1 = -2444.71J$

Therefore,

A) Total heat applied = $Q = 3259.81J$

B) Total change in internal energy

$\delta U = U + U^1 = 0J$

C) Total workdone = $w + w^1$

$= 815 + \frac{0.33*8.314}{\frac{4}{3}-1}*297\\\\= 3259.565J$

D)

$V_3 = 2^3*V_2\\\\= 2^3 * 2(7100*10^{-6})\\\\= 0.1136m^3$

For more information, visit

https://brainly.com/question/14866968