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A cycle tyre bursts suddenly. This represents an

Isothermal process

Isobaric process

Isochoric process

Adiabatic process

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Explanation

The process is very fast, so the gas fails to gain or lose heat. Hence this process in adiabatic

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One mole of helium is adiabatically expanded from its initial state $(P_{i}, V_{i}, T_{i})$ to its final state $(P_{f}, V_{f}, T_{f})$ . The decrease in the internal energy associated with this expansion is equal to

$C_{V} (T_{i} - T_{f})$

$C_{P} (T_{i} - T_{f})$

$\frac{1}{2} (C_{P} + C_{V}) (T i - T_{f})$

$(C_{P} - C_{V}) (T_{i} - T_{f})$

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Explanation

$Δ U = μ C_{V} Δ T = 1 \times C_{V} (T_{f} - T_{i}) = - C_{V} (T_{i} - T_{f})$

⇒ |ΔU| = C_V (T_i – T_f)

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A diatomic gas initially at 18°C is compressed adiabatically to one-eighth of its original volume. The temperature after compression will be

10°C

887°C

668 K

144°C

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Explanation

$T V^{γ - 1} =$ constant

$\Rightarrow T_{2} = T_{1} {(\frac{V_{1}}{V_{2}})}^{γ - 1} = (273 + 18) {(\frac{V}{V / 8})}^{0 .4} = 668 K$

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During an adiabatic process, the pressure of a gas is found to be proportional to the cube of its absolute temperature. The ratio C_p/C_v for the gas is

$\frac{3}{2}$

$\frac{4}{3}$

$\frac{5}{3}$

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Explanation

Given $P \propto T^{3}$ , but we know for an adiabatic process, the pressure $P \propto T^{γ / γ - 1}$

So $\frac{γ}{γ - 1} = 3 \Rightarrow γ = \frac{3}{2} \Rightarrow \frac{C_{P}}{C_{V}} = \frac{3}{2}$

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One mole of an ideal gas at an initial temperature of T K does 6 R joules of work adiabatically. If the ratio of specific heats of this gas at constant pressure and at constant volume is 5/3, the final temperature of gas will be -

(T + 2.4)K

(T – 2.4)K

(T + 4)K

(T – 4)K

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Explanation

$W = \frac{R (T_{i} - T_{f})}{γ - 1}$

⇒ $6 R = \frac{R (T - T_{f})}{(\frac{5}{3} - 1)}$ ⇒ $T_{f} = (T - 4) K .$

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We consider a thermodynamic system. If ΔU represents the increase in its internal energy and W the work done by the system, which of the following statements is true ?

ΔU = –W in an adiabatic process

ΔU = W in an isothermal process

ΔU = –W in an isothermal process

ΔU = W in an adiabatic process

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Explanation

According to the first law of thermodynamics

$Δ Q = Δ U + Δ W$

In adiabatic process $Δ Q = 0$ , hence $Δ U = - Δ W$

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The volume of a gas is reduced adiabatically to $\frac{1}{4}$ of its volume at 27°C, if the value of γ = 1.4, then the new temperature will be -

350 × 4^0.4K

300 × 4^0.4 K

150 × 4^0.4 K

None of these

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Explanation

For adiabatic change $T V^{γ - 1}$ = constant

⇒ $\frac{T_{2}}{T_{1}} = {(\frac{V_{1}}{V_{2}})}^{γ - 1}$ ⇒ $T_{2} = {(\frac{V_{1}}{V_{2}})}^{γ - 1} \times T_{1}$

⇒ $T_{2} = {(\frac{V}{V / 4})}^{1 .4 - 1} \times 300 = 300 \times {(4)}^{0 .4} K$

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For an adiabatic expansion of a perfect gas, the value of $\frac{Δ P}{P}$ is equal to

$- \sqrt{γ} \frac{Δ V}{V}$

$- \frac{Δ V}{V}$

$- γ \frac{Δ V}{V}$

$- γ^{2} \frac{Δ V}{V}$

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Explanation

$P V^{γ} =$ constant : Differentiating both sides

$P γ V^{γ - 1} d V + V^{γ} d P = 0 \Rightarrow \frac{d P}{P} = - γ \frac{d V}{V}$

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A gas expands under constant pressure P from volume V₁ toV₂. The work done by the gas is

$P (V_{2} - V_{1})$

$P (V_{1} - V_{2})$

$P (V_{1}^{γ} - V_{2}^{γ})$

$P \frac{V_{1} V_{2}}{V_{2} - V_{1}}$

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Explanation

Work done $= P Δ V = P (V_{2} - V_{1})$

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When heat in given to a gas in an isobaric process, then

The work is done by the gas

Internal energy of the gas increases

Both (1) and (2)

None from (1) and (2)

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Explanation

When heat is supplied at constant pressure, a part of it goes in the expansion of gas and remaining part is used to increase the temperature of the gas which in turn increases the internal energy.

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