Question

At a given temperature the pressure of a gas reduces to 75% of its initial value and the volume increases by 40% of its initial value. Find this temperature if the initial temperature was – 10°C. Answer- [3.15°C]​

Answer 1

Given :

• Let the initial pressure be P and initial volume be V
• Initial temperature (T) = - 10 °C = 273 - 10 = 263 K
• No of moles (n) remains constant .
• Final pressure = P' =  0.75P = 0.75 P =
• Final volume = V' = V + 0.40 V = 1.4 V

Solution :

Initially,

Ideal gas equation,

PV = nRT

P = pressure (atm)

V = volume (L)

n = moles (mol)

R = Ideal gas constant (0.0821 L .atm/K. mol)

T = Temperature ( K)

⇒ nR = PV/T ...(1)

Now,

⇒ P' V' = n R T'

On rearranging,

⇒ n R= P' V'/T' ...(2)

Equating equation (1) and (2) we get,

⇒ PV / T = P' V'/T'

On rearranging,

⇒ T ' = P' x V' x T/P x V

⇒ T' = 0.75P x 1.4 V x 263/P x V

⇒ T' = 0.75 x 1.4 x 263

⇒ T' = 276.15 K

Conversion from K into °C

⇒ T ' = 276.15 - 273

⇒ T' = 3.15 °C.

∴ The final temperature of the gas is 3.15 °C.

Answer 2

Answer:

Answer :

Iron. The blue colour of aqueous copper sulphate solution can be changed to place green by immersing an iron rod in it.

Explanation :

Out of the given options, all except silver can displace copper from its solution. But, we are also asked that the solution should be changed to a colour of pale-green. Hence, the answer is iron beacuse the colour of iron sulphate solution is pale-green.

When an iron rod is immersed in the copper sulphate solution it displaces copper from its solution and forms iron sulphate solution which gives a pale green colour to the resultant solution.

Reaction :

\sf{CuSO_4(aq)+Fe(s)\longrightarrow{}\underset{\textsf{pale green}}{FeSO_4(aq)}+Cu(s)}CuSO4(aq)+Fe(s)⟶pale greenFeSO4(aq)+Cu(s)

_____________

Additional Information :

Displacement reaction :

A type of chemical reaction in which an atom or group of atoms displace another atom or group of atoms of a compound. Displacement reactions are also called single-displacement reactions.

Examples :

\sf{CuCl_2(aq)+Zn(s)\longrightarrow{}ZnCl_2(aq)+Cu(s)}CuCl2(aq)+Zn(s)⟶ZnCl2(aq)+Cu(s)

\sf{CuSO_4(aq)+Zn(s)\longrightarrow{}ZnSO_4(aq)+Cu(s)}CuSO4(aq)+Zn(s)⟶ZnSO4(aq)+Cu(s)

\sf{CuCl_2(aq)+Pb(s)\longrightarrow{}PbCl_2(aq)+Cu(s)}CuCl2(aq)+Pb(s)⟶PbCl2(aq)+Cu(s)