Question

1 mole of monoatomic ideal gas subjected
to irreversible adiabatic expansion against
constant external pressure of 1 atm starting
from initial pressure of 5 atm and initial
temperature of 300 K till the final pressure is
2 atm. What is the magnitude of work done
(in cal) in the process. (take R = 2 cal/mol K).​

Answer 1

Answer:

The following are the physical properties of metal :

1. The physical properties make them useful for many purposes. For e.g. Copper is used in making electric wires, gold is used to make jewellery, stainless steel is used to make pots, pans, etc
2. Metals react with nonmetals to form ionic bonds. For e.g. Sodium Chloride (NaCl)
3. Metals are a good conductor of electricity which means that they can conduct electricity due to the free moving electrons present in them. Copper is used as wiring as it is a good conductor of electricity.
4. Metals have high melting points and high boiling points as they have strong metallic bonds.
5. All metals are physically lustrous. They have a lustre that makes them shine. Gold is used for making jewellery.
6. Metals are hard, they can’t be broken easily and require a lot of energy and strength to break. Iron is used to make cars, buildings, ships, etc.
7. Metals weigh a lot as they have a high density. Metals are heavy for their size.
8. They don’t have flexibility and have tensile strength. Metals can’t be stretched.

I hope this will be help you.

Answer 2

Answer:

T = 255 K

Explanation:

Topic:

Thermodynamics (Chemistry)

Given:
no. of moles (n) = 1 mole

ideal monoatomic gas ∴ molar heat capacity at const. volume (C_v) = 3R/2

adiabatic process ∴ heat exchanged (Q) = 0

external pressure (p_ext) = 1 atm

initial internal pressure (p_in) = 5 atm → final internal pressure (p_fin) = 2 atm

initial temp (T_in) = 300K                     → final temp (T_fin) = ? = T

gas constant (R) = 2 cal/ mol K

Proof:

ΔU = Q + W

(change in internal energy = heat exchanged + work done)

n × C_v × ΔT = - p_ext × ΔV

(∵ ΔU = n × C_v × ΔT and W = - p_ext × ΔV)

1 × 3R/2 × (T - 300) = - 1 × (V_fin - V_in)

3R/2 × (T - 300) = V_in - V_fin

3R/2 × (T - 300) = (n × R × T_in/P_in) - (n × R × T_fin/P_fin)

(∵ P × V = n × R × T)

3R/2 × (T - 300) = n × R × (T_in/P_in - T_fin/P_fin)

3R/2 × (T - 300) = 1 × R × (300/5 - T/2)

3/2 × (T - 300) = (300/5 - T/2)

3/2 × (T - 300) = (600 - 5T)/10

15 × (T - 300) = (600 - 5T)

15T - 4500 = 600 - 5T

20T = 5100

T = 255K

Any corrections/suggestions for improvement are welcome :)