Question

Half mole of an ideal gas (γ = 5/3) is taken through the cycle abcda, as shown in the figure. Take R=253J K-1 mol-1. (a) Find the temperature of the gas in the states a, b, c and d. (b) Find the amount of heat supplied in the processes ab and bc. (c) Find the amount of heat liberated in the processes cd and da.
Figure

Answer 1

Explanation:

Given:

Number of moles of the gas, $n=0.5 \mathrm{mol}$

$R=\frac{25}{3} \frac{J}{m o l}-K$

$\gamma=\frac{5}{3}$

(a) Temperature at a = Ta

$\mathrm{P}_{\mathrm{a}} \mathrm{V}_{\mathrm{a}}=\mathrm{nRT}_{\mathrm{a}}$

$T_{a}=\frac{P_{a} V_{a}}{n R}=\frac{100 \times 10^{3} \times 5000 \times 10^{-6}}{0.5 \times \frac{25}{3}}=120 K$

Similarly, temperature at b,

$T_{a}=\frac{P_{a} V_{a}}{n R}=\frac{100 \times 10^{3} \times 10000 \times 10^{-6}}{0.5 \times \frac{25}{3}}=240 \mathrm{K}$

Similarly, temperature at c is $480 \mathrm{k}$ and at d  is $240 k$

(b) For process ab,

$\mathrm{d} Q=n c_{p} \mathrm{d} T$

[Since ab is isobaric]

$d Q=\frac{1}{2} \times \frac{R \gamma}{\gamma-1}\left(T_{b}-T_{a}\right)$

$d Q=\frac{1}{2} \times \frac{\frac{(25 \times 5)}{3 \times 3}}{\frac{5}{3}-1} \times(240-120)$

$d Q=\frac{1}{2} \times \frac{125}{9} \times \frac{3}{2} \times(120)$

$d Q=1250 \mathrm{J}$

For line bc, volume is constant. So, it is an isochoric process.

$d Q=d U+d W$

[$d W=0$, isochoric process]

$\mathrm{dQ}=\mathrm{dU}=\mathrm{nC}_{\mathrm{v}} \mathrm{dT}$

$\mathrm{dQ}=\mathrm{nC}_{\mathrm{v}}\left(T_{c}-T_{b}\right)$

$d Q=\frac{1}{2} \times \frac{\frac{25}{3}}{\left[\left(\frac{5}{3}\right)-1\right]} \times(240)$

$d Q=\frac{1}{2} \times \frac{25}{3} \times \frac{3}{2} \times(240)$

$=1500 \mathrm{J}$

(c) Heat liberated in cd (isobaric process),

$\mathrm{dQ}=-\mathrm{nC}_{\mathrm{p}} \mathrm{dT}$

$\mathrm{dQ}=-\frac{1}{2} \times \frac{R \gamma}{\gamma-1}\left(T_{d}-T_{c}\right)$

$d Q=-\frac{1}{2} \times \frac{125}{9} \times \frac{3}{2} \times(240-480)$

$d Q=-\frac{1}{2} \times \frac{125}{9} \times \frac{3}{2} \times(-240)$

$=2500 \mathrm{J}$

Heat liberated in da (isochoric process),

$d Q=d U$

$\mathrm{dQ}=-\mathrm{nC}_{\mathrm{v}} \mathrm{dT}$

$\mathrm{dQ}=-\frac{1}{2} \times \frac{R}{\gamma-1}\left(T_{a}-T_{d}\right)$

$d Q=-\frac{1}{2} \times \frac{25}{2} \times(120-240)$

$d Q=-\frac{1}{2} \times \frac{25}{2} \times(-120)$

$=750 J$

Answer 2

(a) The temperature of the gas in the states a is 120 K

The temperature of the gas in the states b is 240 K

The temperature of the gas in the states c is 480 K

The temperature of the gas in the states d is 240 K

(b) The amount of heat supplied in the processes ab is 1250 J

The amount of heat supplied in the processes bc is 1500 J

(c) The amount of heat liberated in the processes cd is 2500 J

The amount of heat liberated in the processes da is 750 J

Given:

n = 1/2

γ = 5/3

R=253 J/K mol

Explanation:

(a)

The ideal gas is given by the formula:

PV = nRT

Where, P = Pressure; V = Volume; R = Gas constant; T = Temperature

The temperature of the gas in the states a:

$T_a=\frac{P V}{n R}$

On substituting the known values, we get,

$T_a=\frac{5000 \times 10^{-6} \times 100 \times 10^{3}}{\frac{1}{2} \times \frac{25}{3}}$

$\therefore T_a=120 \ K$

The temperature of the gas in the states b:

$T_b=\frac{P V}{n R}$

On substituting the known values, we get,

$T_b=\frac{10000 \times 10^{-6} \times 100 \times 10^{3}}{\frac{1}{2} \times \frac{25}{3}}$

$\therefore T_b=240 \ K$

The temperature of the gas in the states c:

$T_c=\frac{P V}{n R}$

On substituting the known values, we get,

$T_c=\frac{1000 \times 10^{-6} \times 200 \times 10^{3}}{\frac{1}{2} \times \frac{25}{3}}$

$\therefore T_c=480 \ K$

The temperature of the gas in the states d:

$T_d=\frac{P V}{n R}$

On substituting the known values, we get,

$T_d=\frac{5000 \times 10^{-6} \times 200 \times 10^{3}}{\frac{1}{2} \times \frac{25}{3}}$

$\therefore T_d=240 \ K$

(b)

When heat is transferred from a to b, it is an isobaric process.

The heat supplied is given by the formula:

$d Q=n C p d T$

$dQ=\frac{1}{2} \frac{R \gamma}{\gamma-1}(T b-T a)$

On substituting the values, we get,

$dQ=\frac{1}{2} \frac{\left(\frac{25}{3} \times \frac{5}{3}\right)}{\frac{5}{3}-1}(240-120)$

$\therefore dQ=1250 \ \mathrm{J}$

When heat is transferred from b to c, it is an isochoric process.

The heat supplied is given by the formula:

$d Q=n C v d T$

$dQ=\frac{1}{2} \frac{R}{\gamma-1}(T c-T b)$

On substituting the values, we get,

$dQ=\frac{1}{2} \frac{\frac{25}{3}}{\frac{5}{3}-1} \times 240$

$\therefore dQ=1500 \ J$

(c)

cd is an isobaric process.

The heat liberated is given by the formula:

$d Q=-n C p d T$

$dQ=\frac{1}{2} \frac{R \gamma}{\gamma-1}(T d-T c)$

On substituting the values, we get,

$dQ=-\frac{1}{2} \frac{\left(\frac{25}{3} \times \frac{5}{3}\right)}{\frac{5}{3}-1}(-240)$

$\therefore dQ=2500 \ J$

da is an isochoric process.

The heat liberated is given by the formula:

$d Q=-n C v d T$

$dQ=\frac{1}{2} \frac{R}{\gamma-1}(T a-T d)$

On substituting the values, we get,

$dQ=-\frac{1}{2} \frac{\frac{25}{3}}{\frac{5}{3}-1}(120-240)$

$\therefore dQ=750 \ J$