There are two compartments in a rigid fully insulated tank of 14 m3 volume- First compartment contains 8 m3 of a gas c - 1 kJ at 0-7 MPa and 28-C and other contain gas c - 0-7 kJ-kgK 6 m3 at 1MPa and 840-C- If the portion between the gas is removed- what is the final temperature of the gas- Assume ideal gas behaviour

M_A = PVRT= 0.7× 8R× (28+273) = 4215 R m_B = PVRT= 1× 6R× (840+273) = 2371 R ⇒m_Am_B = 3.451 From energy balance: m_Ac_A (T 301) = m_Bc_B(1113 T) 3.451 m ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Physics

Work

There are two...

Question

There are two compartments in a rigid fully insulated tank of $14 m^{3}$ volume. First compartment contains $8 m^{3}$ of a gas (c = 1 kJ\kgK) at 0.7 MPa and $28^{\circ} C$ and other contain gas (c = 0.7 kJ/kgK) $6 m^{3}$ at 1MPa and $840^{\circ} C$ . If the portion between the gas is removed, what is the final temperature of the gas?
(Assume ideal gas behaviour)

{148.4}^{\circ} C

No worries! We‘ve got your back. Try BYJU‘S free classes today!

{164.9}^{\circ} C

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

{178.6}^{\circ} C

No worries! We‘ve got your back. Try BYJU‘S free classes today!

{191.2}^{\circ} C

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is B ${164.9}^{\circ} C$
$m_{A} = \frac{P V}{R T} = \frac{0.7 \times 8}{R \times (28 + 273)} = \frac{4}{215 R}$

$m_{B} = \frac{P V}{R T} = \frac{1 \times 6}{R \times (840 + 273)} = \frac{2}{371 R}$

$\Rightarrow \frac{m_{A}}{m_{B}} = 3.451$

From energy balance:

$m_{A} c_{A} (T - 301) = m_{B} c_{B} (1113 - T)$

$3.451 m_{B} \times 1 (T - 301) = m_{B} \times 0.7 \times (1113 - T)$

$4.93 T - 1483.93 = 1113 - T$

$T = 437.9 K = {164.9}^{\circ} C$

Suggest Corrections

Similar questions

Q. A rigid and insulated tank of 3

m^{3}

volume is divided into two components. One compartment of volume of 2

m^{3}

contains an ideal gas at 0.8314 MPa and 400 K and while the second compartment of volume 1

m^{3}

contains the same gas 8.314 MPa and 500 k. If the partition between the two compartments is ruptured, the final temperature of the gas is:

Q. A rigid and insulated tank of

3 m^{3}

volume is divided into two compartments. One compartment of volume of

2 m^{3}

contains an ideal gas at

0.8314 M P a

and

400 K

and while the second compartment of volume

1 m^{3}

contains the same gas at

8.314 M P a

and

500 K

. If the partition between the two compartments is ruptured, the final temperature of the gas is:

Q. A rigid and insulated tank of

3 m^{3}

volume is divided into two compartments. One compartment of volume of

2 m^{3}

contains an ideal gas at 5 MPa and 400 K and while the second compartment of volume

1 m^{3}

contains the same gas at 50 MPa and 500 K. If the partition between the two compartments is ruptured, the final temperature of the gas is:

Q. A rigid storage tank of 2.5

m^{3}

contains 3 kg argon at 28

^{\circ} C

, heat is then transferred to the argon from a furnace maintained at 1200

^{\circ} C

untill the specific entropy of the argon gas is increased by 0.343 kJ/kgK. what is the percentage increase in absolute temperature of argon gas during the process? Take

C_{v} = 0.312 k J / k g K, R = 0.20 k J / k g K

for argon gas

Q. A container id divided into two compartment. One compartment contains

2

mole of

N_{2}

gas at

1

atm and

300 K

& other compartment contains

H_{2}

gas at the same temperature and pressure. Volume of

H_{2}

compartment is four times the volume of

N_{2}

compartment. [assume no reaction under these condition]
If the container containing

N_{2}

and

H_{2}

are further heated to

1000 K

forming

N H_{3}

with

100 %

yield. Calculate the final total pressure.

Join BYJU'S Learning Program

The correct option is B 164.9∘CmA=PVRT=0.7×8R×(28+273)=4215R mB=PVRT=1×6R×(840+273)=2371R ⇒mAmB=3.451 From energy balance: mAcA(T−301)=mBcB(1113−T) 3.451mB×1(T−301)=mB×0.7×(1113−T) 4.93T−1483.93=1113−T T=437.9 K=164.9∘C