Thermal Properties of Matter Test

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Thermal Properties of Matter Test - 40

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Weekly Quiz Competition

Question 1
1 / -0

A quantity of air $$(\gamma = 1.4)$$ at 27$$^o$$C is compressed suddenly, the temperature of the air system will

A
fall

B
rise

C
remain unchanged

D
first rise and then fall

Solution

Since the gas is compressed suddenly, internal energy increases, so its temperature will increase as work is done on the gas.
Question 2
1 / -0

The temperature to which a gas must be cooled before it can't be liquefied by pressure alone is called its

A
Critical temperature

B
Saturation point

C
Boiling point

D
Freezing point

Solution

$$Answer:-$$ A
In this P-V curve the isotherm having $$T_c$$ is that critical temperature and corresponding to that there is critical pressure $$P_c$$ .

Critical temperature is the temperature of a gas in its critical state, above which it cannot be liquefied by pressure alone.
Critical pressure is the pressure of a gas or vapour in its critical state.
Question 3
1 / -0

Which of the following curve represent variation of density of water with temperature best:

A

B

C

D

Solution

Density of water is maximum at 4℃ and is less on either side of this temperature.
Graph-D represent this best.
Question 4
1 / -0

A 1 liter of an ideal gas at $$27^o C$$ is heated at constant pressure to $$297^o C$$. Its final volume will be:

A
11 litres

B
1.1 litres

C
3.4 litres

D
1.9 litres

Solution

$${V}_{1} = 1000 C. C. {V}_{2} = $$?
$${T}_{1} = 27 C = 300 K ; {T}_{2} = 297 C = 570 K$$
Since the pressure is constant
$$\therefore \displaystyle\frac{{V}_{1}}{{T}_{1}} = \displaystyle\frac{{V}_{2}}{{T}_{2}}$$

$$\therefore \displaystyle\frac{1000}{300} = \displaystyle\frac{{V}_{2}}{570}$$

$$\therefore {V}_{2} = \displaystyle\frac{1000 \times 570}{300} = 1900 c. c.$$
$$= 1.9 litres$$
Question 5
1 / -0

Directions For Questions

Heat and Temperature :

...view full instructions

A manufacture marks the thermometer wrongly.So At $$0^{\circ}C$$ it reads $$10^{\circ}C$$, at $$100^{\circ}C$$ it reads $$85^{\circ}C$$. Then the reading at $$50^{\circ}C$$ will be :

A
$$40^{\circ}C$$

B
$$32.5^{\circ}C$$

C
$$37.5^{\circ}C$$

D
$$42.5^{\circ}C$$

Solution

At $$0^o C$$ there is a difference of $$10^o C$$
At $$100^o C$$ there is a difference of $$15^o C$$
Hence, at $$50^o C$$ there should be a difference of $$12.5^o C$$ i.e. 50-12.5= $$37.5^o C$$
Question 6
1 / -0

Two gases $$A$$ and $$B$$ having the same temperature $$T$$, same pressure $$P$$ and same volume $$V$$ are mixed. If the mixture is at the same volume $$V$$, the pressure of the mixture is

A
$$P$$

B
$${P}/{2}$$

C
$$2P$$

D
$$4P$$

Solution

From Ideal Gas Law,
$$PV=nRT$$
$$\implies$$ Moles of a gas=$$n=\dfrac{PV}{RT}$$
When the mixture is created, the total number of moles is equal to the sum of that of individual gases.
Hence $$N=n_1+n_2$$
$$\implies \dfrac{P'V}{RT}=\dfrac{PV}{RT}+\dfrac{PV}{RT}$$
$$\implies P'=2P$$
Question 7
1 / -0

Real gases obey ideal gas laws more closely at:

A
low pressure and low temperature

B
low pressure and high temperature

C
high pressure and low temperature

D
high pressure and high temperature

Solution

Real gases obey ideal gas laws at low pressure and high temperature because at low pressure the number of molecules per unit volume is less so attractive force between them is negligible. At high temperature the speed of the molecules is very high so collisions becomes elastic.
Question 8
1 / -0

A temperature of $$15^o$$ C on the Fahrenheit scale is ________________.

A
$$59^0$$ F

B
$$27^0$$ F

C
$$-27^0$$ F

D
$$-59^0$$ F

Solution

Solution: There are two different units for the measurement of the temperature. the first one is Celsius and the other one is Fahrenheit.
The relation between the two units is given by,
$$F=\dfrac{9}{5}C+32$$ (1)
where C is the temperature in Celsius and F is the temperature in Fahrenheit.
On Substituting the value of $$C=15^{0}$$ in equation (1) we get,
$$F=\dfrac{9}{5}(15)+32$$
F=$$59^{0}$$
So, temperature in Fahrenheit is $$59^{0}F$$.
Hence, the correct option is (A).
Question 9
1 / -0

The volume of mole of a prefect gas at NTP is ______.

A
22.4 litres

B
2.24 litres

C
100 litres

D
none of these

Solution

At NTP is normal temperature and pressure.
$$\Rightarrow$$ Pressure $$=1\;atm\;;T=293k\;;R=0.08205\dfrac{L\;atm}{mol}$$
$$\Rightarrow V=\dfrac{RT}{P}$$
$$\Rightarrow V=\dfrac{(0.08205)293}{1\;atm}$$
$$\Rightarrow V_{molar}=22.4\;L$$
Hence, the answer is $$22.4\;L.$$
Question 10
1 / -0

The volume of a perfect gas at NTP is

A
22.4 litres

B
2.24 litres

C
100 litres

D
None of these

Solution

At NTP
Pressure $$=1$$ atm
Temperature $$=293k$$
According to ideal gas equation $$PU=nRT$$
$$\Rightarrow V=\dfrac{1\;mol\;(0.08205)293}{1\;atm}$$
$$\Rightarrow (For\;n=1)\;V=22.4\;L$$
Hence, the answer is $$22.4\;L.$$