Q. 34.0( 1 Vote )

# Figure shows a cy

Answer :

**Given:**

Mass of piston(m) = 50 kg

Area(A) = 100 cm^{2} = (100x10^{-4})m^{2} (since 1m = 100 cm)

Acceleration due to gravity, g = 10 ms^{-2}

Atmospheric pressure = 100 kPa

Distance through which it moves = 20 cm

γ = 1.4

**Formula used:**

Therefore, pressure exerted by piston = =

= ((50x10)/(100x10^{-4})) Pa

= 50,000 Pa

Atmospheric pressure = 100 kPa = 1,00,000 Pa.

Therefore, Total pressure(P) = (50,000 + 1,00,000)Pa

=1,50,000Pa

Work done = Pressure x change in volume = P x dV

dV(change in volume) = distance moved by piston x Area

= (20cm x 100cm^{2})

= 2,000 cm^{3} = 2,000 x 10^{-6} m^{3} = 2 x 10^{-3} m^{3}

Therefore, Work = (1,50,000 x 2 x 10^{-3}) J = 300 J

Work done, W= P∆V =n R dT

We get,

Now, We calculate Q:

dQ= nC_{p}dt =

Given: γ = 1.4 = . Also,. Solving these two equations, we get C_{p} = 7R/2, C_{v} = 5R/2.

Hence, dQ = = 1050 J.

Rate this question :

Two vessels A andHC Verma - Concepts of Physics Part 2

4.0 g of helium oHC Verma - Concepts of Physics Part 2

Figure shows two HC Verma - Concepts of Physics Part 2

An amount Q of heHC Verma - Concepts of Physics Part 2

The volume of an HC Verma - Concepts of Physics Part 2

A gas is enclosedHC Verma - Concepts of Physics Part 2

Two samples A andHC Verma - Concepts of Physics Part 2

A sample of an idHC Verma - Concepts of Physics Part 2

An ideal gas (C<sHC Verma - Concepts of Physics Part 2

Half mole of an iHC Verma - Concepts of Physics Part 2