Question

please answer 18-20

Question 18 1 pts Problems 18-20 are based on the following description An insulated rigid tank is divided into two equal parts by a partition. Initially, one part contains 10 kg of an ideal gas with molar mass M = 16 kg/kmol, a pressure of 330 kPa and a temperature of 50 oC. The other part is evacuated. The partition is now removed and the gas fills the entire tank (R.- 8.31447 k/MOLK) The expansion work in ki is closer to: 1678 25 o 267 820 3215 512

Answer 1

Data provided:

mass of ideal gas(m) =10kgs

M=16kg/kmol

Pressure (P) = 320KPa

Temperature(T) =50 degree Celsius

As show in the attached image,

the other empty half of the tank is opened up and ideal gas is allowed to flow ito it.

1). The expansion work:

0 KJ

Since other half of the chamber is completly vaccum, there is no resisting force in the system. Therefore, the work done by expanding the gas is zero.

2). The final temperature of the ideal gas:

323K or same(50 degree Celsius)

As the tank remains completely insulated, the heat transfer with the environment is zero. Therefore Q=0. Also, as mentioned before the Work done by expansion is zero.

The chambers are at same elevation, therefore we can assume that Kinetic and potential energy change is also zero.

Writing down the energy balance equation,

$Q^{0}-W^{}0= \Delta U +\Delta K.E + \Delta P.E$

Therefore change in internal energy $\Delta U =0$ . Since internal energy solely depends on temperature, there is no change in temperature because internal energy change is zero.

3). Change in entropy of gas during the process.

Entropy change for an ideal gas is given as

$\Delta S = NRln(V_2/V_1)$

V_2= 2*V_1 (Because the tank volume become twice of the initial volume as the partition is removed)

N= no. of moles = m/M = 10/16 = 0.625 * 8.31447 * ln(2)

= 3.6019 KJ/kg.K close to 3.6KJ/Kg.K

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