Q. 94.0( 2 Votes )

# An ideal gas is t

Answer :

**Given:**

P = kV …

Where,

P = pressure,

V = volume,

k = constant.

**Formula used:**

Equation of state of ideal gas:

PV = nRT = constant … (ii)

Where,

n is the number of moles of the gas,

R is the gas constant,

T is the temperature,

P = pressure, V^{2}

T = temperature.

From (i), multiplying by dV on both sides:

PdV = kVdV.

Integrating from V = V1 to V2, we get

= with lower limit V_{1} and upper limit V_{2}

=

Now, we know, PV = nRT - equation of state,

Where P = pressure, V = volume, n = number of moles, R = universal gas constant, T = temperature

Hence we can write, V_{1} = nRT_{1}/P_{1}. Since P_{1} = kV_{1}, this becomes:

kV_{1}^{2} = nRT_{1}. Similarly, KV_{2}^{2} = nRT_{2},

P_{1}, V_{1}, T_{1} - Pressure, volume, temperature of first gas

P_{2}, V_{2}, T_{2} - Pressure, volume, temperature of second gas

Therefore, subsituting, the above equation becomes:

= = … (iii)

Now, => (since P = kV) … (iv)

But Q = U + ഽPdV (first law of thermodynamics), where Q = heat, U = change in internal energy, W = total work done = ഽPdV

=> nCdT = nC_{v}dT + (nR/2)dT

(since Q = nCdT and U = nC_{v}dT)

=> C = C_{v} + nR/2 (proved),

Where

n = number of moles,

C = specific heat capacity,

C_{v} = specific heat capacity at constant volume,

R = universal gas constant,

dT = rise in temperature.

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