Question

A gas is enclosed in a cylindrical can fitted with a piston. The walls of the can and the piston are adiabatic. The initial pressure, volume and temperature of the gas are 100 kPa, 400 cm3 and 300 K, respectively. The ratio of the specific heat capacities of the gas, Cp / Cv = 1.5. Find the pressure and the temperature of the gas if it is (a) suddenly compressed (b) slowly compressed to 100 cm3.

Answer 1

Answer:

Explanation:

Given data:

Initial pressure "P1" = 100 KPa = 105 Pa

Volume "V1"  = 400 cm3 = 400 × 10–6 m3

Temperature "T1" = 300 k

Ratio of the specific heat capacities "γ" = Cp/Cv = 1.5

• (a) Suddenly compressed to V2 = 100 cm3​

P1V1^γ  = P2V2^γ

10^5(400)^1.5 =  P2(100)^1.5

P2 = 10^5 x (4) ^1.4 = 800 KPa

T1V1^γ-1  = T2V2^γ-1  

300 x (400)^1.5-1 = T2 x(100)^1.5-1

T2 = 300 x 20 /10 = 600 K

• (b) slowly compressed to 100 cm3.

Even if the container is slowly compressed the walls are adiabatic therefore the  heat transferred is 0. Thus the values remain same i.e. P2 = 800 KPa, T2 = 600 K.

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Answer 2

(a) The pressure and temperature of the gas when suddenly compressed is 800 kPa and 600 K

(b) The pressure and temperature of the gas when slowly compressed to 100 cm³ is 800 kPa and 600 K

Explanation:

Given Data

Initial Pressure of the gas , $P_1 = 100 kPa$

Initial Volume of the gas ,  $V_1 = 400 cm^{3}$

                                                = $400 \times 10^{-6} \mathrm{m}^{3}$                                      

Initial temperature of the gas $T_1 = 300 K$

$\gamma=\frac{C_{p}}{C_{v}}=1.5$

(a) Suddenly gas is compressed

The gas is compressed to volume suddenly, $V_{2}=100 \mathrm{cm}^{3}$

This, then, is an adiabatic method.

For a process of adiabatic,

$\mathrm{P}_{1} \mathrm{V}_{1}^{\gamma}=\mathrm{P}_{2} \mathrm{V}_{2}^{\gamma}$

$10^{5} \times(400)^{1.5}=\mathrm{P}_{2} 100^{1.5}$

$\mathrm{P}_{2}=10^{5} \times 40^{1.5}$

$P_2 =10^{5} \times 8$

     =100000×8

    $=800 \times 10^{3}$

     =800 kPa

$P_{2}=800 \mathrm{kPa}$

$T_{1} V^{-1}=T_{2} V_{2}^{v-1}$

$300 \times(400)^{1.5-1}=T_{2}(100)^{1.5-1}$

$300 \times(400)^{0.5}=T_{2}(100)^{0.5}$

$T_{2}=600 \mathrm{K}$

(b) Slowly compressed to 100 cm³

If the container is squeezed gradually the heat transfer is zero, even if the walls are adiabatic.  

The values, then, remains the same. Accordingly,

$P_{2}=800 \mathrm{kPa}$

$T_{2}=600 \mathrm{K}$

Therefore when a gas is enclosed in cylindrical can fitted with piston with initial pressure, volume and temperature as 100 kPa, 400 cm³, and 300 K, then the pressure and temperature at compressed state is 800 kPa and 600 K and also remains the same pressure and temperature as 800 kPa and 600 K when slowly compressed to 100 cm³