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1st Law of Thermodynamics

Two moles of ...

Question

Two moles of a monoatomic ideal gas $[U = \frac{3}{2} n R T]$ is enclosed in an adiabatic vertical cylinder fitted with a smooth light adiabatic piston. The piston is connected to a vertical spring of spring constant 200 N/m as shown in figure. The area of cross-section of the cylinder is $20.0 c m^{2}$ . Initially, the spring is at its natural length and temperature of the gas is 300K. The atmospheric pressure is 100 kPa. The gas is heated slowly for some time by means of an electric heater so as to move the piston up through 10 cm.

The quantity of heat supplied by the heater is:

793 J

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0.793 J

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79.3 J

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7.93 J

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Solution

The correct option is A 793 J
The internal energy is : $U = (\frac{3}{2}) n R T$

$∴ d U = (\frac{3}{2}) n R Δ T$

$= 1.5 \times (2.0 m o l) \times (8.3 J / m o l - K) \times (31 K) = 772 J$

Hence, according to first law of thermodynamics,
$Δ Q + Δ U + Δ W = 772 J + 21 J = 793 J$

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Similar questions

Q. Two moles of a monoatomic ideal gas

[U = \frac{3}{2} n R T]

is enclosed in an adiabatic vertical cylinder fitted with a smooth light adiabatic piston. The piston is connected to a vertical spring of spring constant 200 N/m as shown in figure. The area of cross-section of the cylinder is

20.0 c m^{2}

. Initially, the spring is at its natural length and temperature of the gas is 300K. The atmospheric pressure is 100 kPa. The gas is heated slowly for some time by means of an electric heater so as to move the piston up through 10 cm.

The work done by the gas in the whole process is

2

moles of a monatomic ideal gas is enclosed in an adiabatic fixed vertical cylinder fitted with a smooth, light and adiabatic piston. The piston is connected to a vertical spring of spring constant

200 N/m

as shown in figure. The area of cross-section of the cylinder is

20 {cm}^{2}

. Initially, the spring is at its natural length and temperature of the gas is

300 K

. The atmospheric pressure is

100 kPa

. The gas is heated slowly for some time by means of an electric heater so as to move the piston up (slowly) by

10 cm

. Then:

Q. An ideal monoatomic gas is confined in a cylinder by a spring-loaded piston of area of cross-section

8 \times 10^{- 3} m^{2}

. Initially, the gas is at 300 K and occupies a volume of

2.4 \times 10^{- 3} m^{3}

and the spring is in its relaxed state as shown in figure. The gas is heated by a small electric heater until the piston moves out slowly by 0.1 m. The force constant of the spring is 8000 N

m^{- 1}

, atmospheric pressure is

1 \times 10^{5} N m^{- 2}

. Find the heat supplied by the heater. The piston is massless and no friction exits between the piston and the cylinder. Neglect the heat loss through the lead wires of heater. The heat capacity of the heater coil is negligible.

An ideal monoatomic gas is confined in a cylinder by a spring loaded piston of cross section

8.0 \times 10^{- 3} m^{2}

. Initially the gas is at

300 K

and occupies a volume of

2.4 \times 10^{- 3} m^{3}

and the spring is in its relaxed state as shown in figure. The gas is heated by a small heater until the piston moves out slowly by

0.1 m

. The force constant of the spring is 8000 N/m and the atmospheric pressure is

1 \times 10^{5} N / m^{2}

. The cylinder and the piston are thermally insulated. The piston and the spring are massless and there is no friction between the piston and the cylinder. The final temperature of the gas will be :

(Neglect the heat loss through the lead wires of the heater. The heat capacity of the heater coil is also negligible)

Q. An ideal monoatomic gas is confined in a cylinder by a spring loaded piston of cross sectional area

4 \times 10^{- 3} m^{2}

. Initially the gas is at

27^{\circ} C

and volume of gas is

1.2 \times 10^{- 3} m^{3}

, system is in equilibrium and spring is in natural state. The sample is heated by a heater until piston moves out slowly by

5 cm

. Calculate final temperature of the gas and heat supplied by heater. Take

k

(spring constant) =

8000 N / m

, mass of piston =

20 kg

(Assume there is no loss of heat through coils and walls, take

P_{a} t m = 10^{5} Pa

)

Final temperature of the gas (in

K

):
Heat supplied by heater (in

J

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The correct option is A 793 JThe internal energy is : U=(32)nRT ∴dU=(32)nRΔT =1.5×(2.0mol)×(8.3J/mol−K)×(31K)=772 J Hence, according to first law of thermodynamics, ΔQ+ΔU+ΔW=772 J+21 J=793 J

The correct option is A 793 J
The internal energy is : $U = (\frac{3}{2}) n R T$

$∴ d U = (\frac{3}{2}) n R Δ T$

$= 1.5 \times (2.0 m o l) \times (8.3 J / m o l - K) \times (31 K) = 772 J$

Hence, according to first law of thermodynamics,
$Δ Q + Δ U + Δ W = 772 J + 21 J = 793 J$