Question

One mole of an ideal gas is compressed from 500 cm3 against constant pressure of 1.216 × 105 pa. The work involved in the process is 36.50 j. Calculate the final volume.

Answer 1

Explanation:

Initial volume of the gas = V_1=500 cm^3=500\times 10^{-6} m^3=0.0005 m^3V

1

=500cm3

=500×10^-6 m3

=0.0005m3

Final volume of gas = V_2V

2

External pressure p_{ext}= 1.216\times 20^{-5} pa=1.216\times 10^{-5} n/m^2p

ext

=1.216×20

−5

pa=1.216×10

−5

n/m

2

The work involved in the system = w = 36.50 J = 36.50 N/m

w=-p_{ext}\times (V_2-V_1)w=−p

ext

×(V

2

−V

1

)

36.50 N m=-1.216\times 10^{5} N/m^2\times (V_2-0.0005 m^3)36.50Nm=−1.216×10

5

N/m

2

×(V

2

−0.0005m

3

)

V_2=200\times 10^{-5] m^3=200 cm^3

The final volume of the gas is 200 cm^3200cm

3

.