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Byju's Answer

Standard XII

Chemistry

Ideal Gas Equation

4 g argon Ato...

Question

4 g argon (Atomic mass=40) in a bulb is at a temperature of T K has a pressure P atm. When the bulb was placed in hot bath at a temperature $50^{0} C$ more than the first one, 0.8 g of gas had to be removed to get the original pressure. T is equal to :

510 k

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200 k

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100 k

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73 k

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Solution

The correct option is D 200 k
$\begin{matrix} P V = n R T P = \frac{n R T}{V} S o, \frac{4}{40 V} R T = \frac{3.2}{40 V} R (T + 50) \Rightarrow 40 T = 32 (T + 50) 40 T - 32 T = 50 \times 32 T = \frac{50 \times 32}{8} \Rightarrow 200 K \end{matrix}$
Option B is correct.

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Similar questions

4 g

of argon (Atomic mass = 40) in a bulb at a temperature of

T K

has a pressure

P

atm. When the bulb was placed in a hot bath at a temperature

50^{\circ} C

more that the first one,

0.8 g

of the gas had to be removed to get the original pressure.

T

is equal to:

4

g argon (Atomic mass

= 40

) in a bulb at a temperature of TK has a pressure P atm. When the bulb was placed in hot bath at a temperature

50^{o}

C more than the first one,

0.8

g of gas had to be removed to get the original pressure. T is equal to :

4 g

of argon (Atomic mass = 40) in a bulb at a temperature of

T K

has a pressure

P

atm. When the bulb was placed in a hot bath at a temperature

50^{\circ} C

more that the first one,

0.8 g

of the gas had to be removed to get the original pressure.

T

is equal to:

Q. At a temperature,

T i n K

, the pressure of

4 g

argon in a bulb is

P

. The bulb is put in a bath having a temperature higher by

50 K

than the first one.

0.8 g

of argon gas had to be removed to maintain original pressure. The temperature

T

is equal to :

Q. At a temperature '

T

K

, the pressure of 4.0

g

of argon in a bulb is

p

. The bulb is put in a bath having temperature higher by 50

K

than the first one. 0.8

g

of argon gas had to be removed to maintain original pressure. The temperature

T

is equal to:

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4 g argon (Atomic mass=40) in a bulb is at a temperature of T K has a pressure P atm. When the bulb was placed in hot bath at a temperature 500C more than the first one, 0.8 g of gas had to be removed to get the original pressure. T is equal to :

The correct option is D 200 kPV=nRTP=nRTVSo,440VRT=3.240VR(T+50)⇒40T=32(T+50)40T−32T=50×32T=50×328⇒200KOption B is correct.

4 g argon (Atomic mass=40) in a bulb is at a temperature of T K has a pressure P atm. When the bulb was placed in hot bath at a temperature $50^{0} C$ more than the first one, 0.8 g of gas had to be removed to get the original pressure. T is equal to :

The correct option is D 200 k
$\begin{matrix} P V = n R T P = \frac{n R T}{V} S o, \frac{4}{40 V} R T = \frac{3.2}{40 V} R (T + 50) \Rightarrow 40 T = 32 (T + 50) 40 T - 32 T = 50 \times 32 T = \frac{50 \times 32}{8} \Rightarrow 200 K \end{matrix}$
Option B is correct.