Physics, asked by adarsh8597, 4 months ago

One kg of air in a closed system, initially at 5°C
and occupying 0.3 m3 volume, undergoes a
constant pressure heating process to 100°C.
There is no work other than pdv work. Find
i) The work done during the process.
ii) The heat transferred.
E

Answers

Answered by Anonymous
0

Answer:

Science, any system of knowledge that is concerned with the physical world and its phenomena and that entails unbiased observations and systematic experimentation. In general, a science involves a pursuit of knowledge covering general truths or the operations of fundamental laws.

Answered by archanajhaasl
0

Answer:

(i) The work done during the process is 243.4 Joules.

(ii)  The heat transferred is 243.4 Joules.

Explanation:

From the ideal gas equation we have,

\mathrm{PV=nRT}         (1)

Where,

P=pressure at which the gas is present

V=volume occupied by the gas

n=number of moles of the gas

R=universal gas constant=0.082 L⋅atm⋅K⁻¹⋅mol⁻¹

T=temperature in kelvin

Equation (1) can also be written as,

\mathrm{P\Delta V=nR\Delta T}        (2)

(i) The work done during the process

And the work done is calculated as,

\mathrm{W=P\Delta V}          (3)         (work done at constant pressure)

Using equation (2) in equation (3) we get;

\mathrm{W=nR\Delta T}            (4)

From the question we have,

The mass of the air=1 kg=1×10³grams

The volume occupied by the gas(V)=0.3 m³

The initial temperature(T₁)=5°C=278K

The final temperature(T₂)=100°C=373K

Inserting all the required values in equation (4) we get;

\mathrm{W=\frac{1\times 10^3}{32} \times 0.082\times (373-278)}

\mathrm{W=243.4\ Joules}         (5)

The work done during the process is 243.4 Joules.

(ii) The heat transferred

From the first law of thermodynamics we have,

\mathrm{Q=\Delta U+W}             (6)

Where,

Q=heat transferred

ΔU=change in internal energy

W=work done during the process

Since the pressure is constant so internal energy will be constant.

So, equation (6) can be written as,

\mathrm{Q=0+W}

\mathrm{Q=W=243.4\ Joules }       (7)

So, the heat transferred is 243.4 Joules.

#SPJ2

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