In terms of Charles' law, explain why - 273 \displaystyle ^{\circ}C is the lowest possible temperature.

At 0 \ K or \displaystyle -273^oC , the volume of the gas becomes equal to zero . The gas ceases to exists. Hence, \displaystyle -273^oC is the lowest possible temperature.

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>>Class 11
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>>Thermal Properties of Matter
>> $Heat and Temperature$

Heat and Temperature

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Heat and Temperature

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Q: A circular metal loop is heated at point O as shown in above Figure.i) In which direction would heat flow in the loop?ii) In which order the pins at points P, Q and R fixed with the help of wax fall if points O, P, Q and R are equidistant from each other?

According to the given condition.i) The heat will flow in the direction from O to P and O to R.ii) Firstly the pins at R and P will fail concurrently followed by the pin at Q.

Q: 1 mole of H_2SO_4 is mixed with 2 moles of NaOH. The heat evolved will be:

{H}_{2}S{O}_{4} + 2 NaOH \longrightarrow {Na}_{2}S{O}_{4} + 2 {H}_{2}O According to reaction, 1 mole {H}_{2}S{O}_{4} completely neutralised by 2 mole of NaOH .As we know,Gram equivalent = no. of moles \times Valency factorValency factor of {H}_{2}S{O}_{4} = 2 Therefore,Gram equivalent of {H}_{2}S{O}_{4} = 1 \times 2 = 2 As we know,Heat evolved in the neutralisation of 1 gm eq. of strong acid = 57.3 \; kJ Heat evolved in neutralisation of 2 gm eq. of strong acid = \left( 57.3 \times 2 \right) kJ Hence the heat evolved will be 2 \times 57.3 \; kJ .Hence, the correct option is \text{B}

Q: The value of absolute zero temperature in Fahrenheit scale is

Absolute temperature in Celsius is -273.15{{\,}^{o}}C F=\dfrac{9}{5}C+32 F=\dfrac{9}{5}\times (-273.15)+32=-459.67{{\,}^{o}}F F\cong -460{{\,}^{o}}F Hence, absolute temperature in Fahrenheit

Q: Figure shows two flasks connected to each other. The volume of the flask 1 is twice that of flask 2. The system is filled with an ideal gas at temperature 100 K and 200 K respectively in the flasks. If the mass of the gas in 1 m, then what is the mass of the gas in flask 2 in equilibrium?

At equilibrium pressure in both the flasks must be same.Using ideal gas equation; \dfrac { { n }_{ 1 }R{ T }_{ 1 } }{ { V }_{ 1 } } =\dfrac { { n }_{ 2 }{ RT }_{ 2 } }{ { V }_{ 2 } } \dfrac { { m }_{ 1 }\times100K }{ { 2V }_{ 2 } } =\dfrac { { m }_{ 2 }\times200K }{ { V }_{ 2 } } { m }_{ 2 }=\dfrac { m }{ 4 } So, the correct option is C

Q: Absolute zero can be defined as the temperature at which:

A gas ceases to exist at absolute zero. It is the lowest temperature which can be attained, where volume becomes zero.

Q: For the reaction : 2A(g)+3B\longrightarrow 4C(G) _D (l); \Delta H-=-20 kJ/mole Find the heat evolved when 0.4 mole of A reacts with an excess of B in a closed rigid container. The temperature is constant at 300 K . [R=8.3 J/K mole]

2 {A}_{\left( g \right)} + 3 {B}_{\left( g \right)} \longrightarrow 4 {C}_{\left( g \right)} + {D}_{\left( l \right)} No. of moles of A = 0.4 No. of moles of C when 2 moles of A reacts = 4 \text{ moles} No. of moles of C when 0.4 moles of A reacts = 0.8 \text{ moles} No. of moles of B = 3 (As B is in excess)Total moles of reactant = 0.4 + 3 = 3.4 Total moles of product = 0.8 {\Delta{n}}_{g} = {n}_{P} - {n}_{R} = 0.8 - 3.8 = -3 \Delta{H} = -20 \; {KJ}/{mol}= -20000\; {J}/{mol} Temperature \left( T \right) = 300 K \Delta{H} = {q}_{V} + {\Delta{n}}_{g}RT \Rightarrow {q}_{V} = \Delta{H} - {\Delta{n}}_{g} RT \Rightarrow {q}_{V} = -20000 - \left( -3 \times 8.3 \times 300 \right) = -20000 + 7470 = -12530 \; {J}/{mol} = -12.53 \; {KJ}/{mol} Hence the required answer is -12.53 \; {KJ}/{mol} .

Q: If the heat change at constant volume for decomposition of silver oxide is 80.25 kJ. What will be the heat change at constant pressure?

Solution:- (B) > 80.25 \; KJ Heat change at constant volume for the decomposition of silver oxide = 80.25 \; KJ {{Ag}_{2}O}_{\left( s \right)} \longrightarrow 2 {Ag}_{\left( s \right)} + \cfrac{1}{2} {{O}_{2}}_{\left( g \right)} From the above reaction, \Delta{{n}_{g}} > 0 From first law of thermodynamics, \Delta{H} = \Delta{E} + \Delta{{n}_{g}} RT \because \Delta{{n}_{g}} > 0 \therefore \Delta{H} > \Delta{E} Hence \Delta{H} > 80.25 \; KJ

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Important Questions

A circular metal loop is heated at point $O$ as shown in above Figure.
i) In which direction would heat flow in the loop?
ii) In which order the pins at points P, Q and R fixed with the help of wax fall if points O, P, Q and R are equidistant from each other?

Revise with Concepts

Heat and Temperature

Learn with Videos

Quick Summary With Stories

Introduction to Heat and Temperarture

Measurement of Temperature

Numericals Problem on Measuring Temperature

Important Questions

A circular metal loop is heated at point O as shown in above Figure. i) In which direction would heat flow in the loop? ii) In which order the pins at points P, Q and R fixed with the help of wax fall if points O, P, Q and R are equidistant from each other?

In terms of Charles' law, explain why - 273 ∘C is the lowest possible temperature.

1 mole of H2​SO4​ is mixed with 2 moles of NaOH. The heat evolved will be:

The value of absolute zero temperature in Fahrenheit scale is

Figure shows two flasks connected to each other. The volume of the flask 1 is twice that of flask 2. The system is filled with an ideal gas at temperature 100 K and 200 K respectively in the flasks. If the mass of the gas in 1 m, then what is the mass of the gas in flask 2 in equilibrium?

A wooden spoon is dipped in a cup of ice cream. It's another end

Absolute zero can be defined as the temperature at which:

For the reaction : 2A(g)+3B⟶4C(G)D​(l);ΔH−=−20kJ/mole Find the heat evolved when 0.4 mole of A reacts with an excess of B in a closed rigid container. The temperature is constant at 300K. [R=8.3 J/K mole]

5g of ice at 0oC is mixed with 5 g of steam at 100oC, what is the final temperature?

If the heat change at constant volume for decomposition of silver oxide is 80.25 kJ. What will be the heat change at constant pressure?

A circular metal loop is heated at point $O$ as shown in above Figure.
i) In which direction would heat flow in the loop?
ii) In which order the pins at points P, Q and R fixed with the help of wax fall if points O, P, Q and R are equidistant from each other?

In terms of Charles' law, explain why - 273 $^{\circ} C$ is the lowest possible temperature.

$1$ mole of $H_{2} S O_{4}$ is mixed with $2$ moles of NaOH. The heat evolved will be:

For the reaction : $2 A (g) + 3 B ⟶ 4 C (G)_{D} (l); Δ H - = - 20 k J / m o l e$
Find the heat evolved when $0.4$ mole of A reacts with an excess of B in a closed rigid container. The temperature is constant at $300 K$ . [R=8.3 J/K mole]

5g of ice at $0^{o} C$ is mixed with 5 g of steam at $10 0^{o} C,$ what is the final temperature?