MCQ

MCQ 1011 Mark

A thermo-dynamical system is changed from state $\left(P_1, V_1\right)$ to $\left(P_2, V_2\right)$ by two different process. The quantity which will remain same will be

A
$\Delta Q$
B
$\Delta W$
C
$\Delta Q+\Delta W$
✓
$\Delta Q-\Delta W$

Answer

Correct option: D.

$\Delta Q-\Delta W$

(d) Change in internal energy does not depend upon path so $\Delta U=\Delta Q-\Delta W$ remain constant.

View full question & answer→

MCQ 1021 Mark

When an ideal gas $(\gamma=5 / 3)$ is heated under constant pressure, then what percentage of given heat energy will be utilised in doing external work

✓
$40 \%$
B
$30 \%$
C
$60 \%$
D
$20 \%$

Answer

Correct option: A.

$40 \%$

(a)$\Delta Q=\Delta U+\Delta W \Rightarrow \frac{\Delta W}{\Delta Q}=1\frac{\Delta U}{\Delta Q}=1-\frac{n C_V d T}{n C_P d T} $
$\Rightarrow \frac{\Delta W}{\Delta Q}=1\frac{C_V}{C_P}=1-\frac{3}{5}=\frac{2}{5}=0.4$

View full question & answer→

MCQ 1031 Mark

Two moles of an ideal monoatomic gas at $27^{\circ} \mathrm{C}$ occupies a volume of $v$. If the gas is expanded adiabatically to the volume $2 \mathrm{~V}$, then the work done by the gas will be $[\gamma=5 / 3, R=8.31 \mathrm{~J} / \mathrm{mol} \mathrm{K}]$

A
$-2767.23 \mathrm{~J}$
✓
$2767.23 \mathrm{~J}$
C
$2500 \mathrm{~J}$
D
$-2500 \mathrm{~J}$

Answer

Correct option: B.

$2767.23 \mathrm{~J}$

$W =\frac{\mu R\left(T_1-T_2\right)}{(\gamma-1)}=\frac{\mu R T_1}{(\gamma-1)}\left[1-\frac{T_2}{T_1}\right] =\frac{\mu RT_1}{(\gamma-1} \left[1\left(\frac{V_1}{V_2}\right)^{\gamma-1}\right] v$
$ =\frac{2 \times 8.31 \times 300}{\left(\frac{5}{3}-1\right)}\left[1-\left(\frac{1}{2}\right)^{\frac{5}{3}-1}\right]=+2767.23 \mathrm{~J}$

View full question & answer→

MCQ 1041 Mark

A thermally insulated container is divided into two parts by a screen. In one part the pressure and temperature are $P$ and $T$ for an ideal gas filled. In the second part it is vacuum. If now a small hole is created in the screen, then the temperature of the gas will

A
Decrease
B
Increase
✓
Remain same
D
None of the above

Answer

Correct option: C.

Remain same

(c) This is the case of free expansion of gas. In free expansion $\Delta U=0 \Rightarrow$ Temp. remains same.

View full question & answer→

MCQ 1051 Mark

A thermo-dynamical system is changed from state $\left(P_1, V_1\right)$ to $\left(P_2, V_2\right)$ by two different process. The quantity which will remain same will be

A
$\Delta Q$
B
$\Delta W$
C
$\Delta Q+\Delta W$
✓
$\Delta Q-\Delta W$

Answer

Correct option: D.

$\Delta Q-\Delta W$

(d) Change in internal energy does not depend upon path so $\Delta U=\Delta Q-\Delta W$ remain constant.

View full question & answer→

MCQ 1061 Mark

When an ideal gas $(\gamma=5 / 3)$ is heated under constant pressure, then what percentage of given heat energy will be utilised in doing external work

✓
$40 \%$
B
$30 \%$
C
$60 \%$
D
$20 \%$

Answer

Correct option: A.

$40 \%$

(a)$\Delta Q=\Delta U+\Delta W \Rightarrow \frac{\Delta W}{\Delta Q}=1\frac{\Delta U}{\Delta Q}=1-\frac{n C_V d T}{n C_P d T} $
$\Rightarrow \frac{\Delta W}{\Delta Q}=1\frac{C_V}{C_P}=1-\frac{3}{5}=\frac{2}{5}=0.4$

View full question & answer→

MCQ 1071 Mark

Two moles of an ideal monoatomic gas at $27^{\circ} \mathrm{C}$ occupies a volume of $v$. If the gas is expanded adiabatically to the volume $2 \mathrm{~V}$, then the work done by the gas will be $[\gamma=5 / 3, R=8.31 \mathrm{~J} / \mathrm{mol} \mathrm{K}]$

A
$-2767.23 \mathrm{~J}$
✓
$2767.23 \mathrm{~J}$
C
$2500 \mathrm{~J}$
D
$-2500 \mathrm{~J}$

Answer

Correct option: B.

$2767.23 \mathrm{~J}$

(b)$\begin{aligned}W & =\frac{\mu R\left(T_1-T_2\right)}{(\gamma-1)}=\frac{\mu R T_1}{(\gamma-1)}\left[1-\frac{T_2}{T_1}\right] \\&=\frac{\muRT_1{(\gamma-1\left[1\left(\frac{V_1}{V_2}\right)^{\gamma-1}\right] \\& =\frac{2 \times 8.31 \times 300{\left(\frac{5{3}-1\right)}\left[1-\left(\frac{1}{2}\right)^{\frac{5}{3}-1}\right]=+2767.23 \mathrm{~J}\end{aligned}$

View full question & answer→

MCQ 1081 Mark

A thermally insulated container is divided into two parts by a screen. In one part the pressure and temperature are $P$ and $T$ for an ideal gas filled. In the second part it is vacuum. If now a small hole is created in the screen, then the temperature of the gas will

A
Decrease
B
Increase
✓
Remain same
D
None of the above

Answer

Correct option: C.

Remain same

(c) This is the case of free expansion of gas. In free expansion $\Delta U=0 \Rightarrow$ Temp. remains same.

View full question & answer→

MCQ 1091 Mark

A thermo-dynamical system is changed from state $\left(P_1, V_1\right)$ to $\left(P_2, V_2\right)$ by two different process. The quantity which will remain same will be

A
$\Delta Q$
B
$\Delta W$
C
$\Delta Q+\Delta W$
✓
$\Delta Q-\Delta W$

Answer

Correct option: D.

$\Delta Q-\Delta W$

(d) Change in internal energy does not depend upon path so $\Delta U=\Delta Q-\Delta W$ remain constant.

View full question & answer→

MCQ 1101 Mark

When an ideal gas $(\gamma=5 / 3)$ is heated under constant pressure, then what percentage of given heat energy will be utilised in doing external work

✓
$40 \%$
B
$30 \%$
C
$60 \%$
D
$20 \%$

Answer

Correct option: A.

$40 \%$

(a)$\Delta Q=\Delta U+\Delta W \Rightarrow \frac{\Delta W}{\Delta Q}=1\frac{\Delta U}{\Delta Q}=1-\frac{n C_V d T}{n C_P d T} $
$\Rightarrow \frac{\Delta W}{\Delta Q}=1\frac{C_V}{C_P}=1-\frac{3}{5}=\frac{2}{5}=0.4$

View full question & answer→

MCQ 1111 Mark

Two moles of an ideal monoatomic gas at $27^{\circ} \mathrm{C}$ occupies a volume of $v$. If the gas is expanded adiabatically to the volume $2 \mathrm{~V}$, then the work done by the gas will be $[\gamma=5 / 3, R=8.31 \mathrm{~J} / \mathrm{mol} \mathrm{K}]$

A
$-2767.23 \mathrm{~J}$
✓
$2767.23 \mathrm{~J}$
C
$2500 \mathrm{~J}$
D
$-2500 \mathrm{~J}$

Answer

Correct option: B.

$2767.23 \mathrm{~J}$

$W =\frac{\mu R\left(T_1-T_2\right)}{(\gamma-1)}=\frac{\mu R T_1}{(\gamma-1)}\left[1-\frac{T_2}{T_1}\right] =\frac{\mu RT_1}{(\gamma-1} \left[1\left(\frac{V_1}{V_2}\right)^{\gamma-1}\right] v$
$ =\frac{2 \times 8.31 \times 300}{\left(\frac{5}{3}-1\right)}\left[1-\left(\frac{1}{2}\right)^{\frac{5}{3}-1}\right]=+2767.23 \mathrm{~J}$

View full question & answer→

MCQ 1121 Mark

If $C_V=4.96 \mathrm{cal} /$ mole $K$, then increase in internal energy when temperature of $2$ moles of this gas is increased from $340 K$ to $342 K$

A
$27.80 \mathrm{cal}$
✓
$19.84 \mathrm{cal}$
C
$13.90 \mathrm{cal}$
D
$9.92 \mathrm{cal}$

Answer

Correct option: B.

$19.84 \mathrm{cal}$

$\Delta U=\mu C_V \Delta T$
$=2 \times 4.96 \times(342-340)$
$=19.84 \mathrm{cal}$

View full question & answer→

MCQ 1131 Mark

If $C_V=4.96 \mathrm{cal} /$ mole $K$, then increase in internal energy when temperature of $2$ moles of this gas is increased from $340 K$ to $342 K$

A
$27.80 \mathrm{cal}$
✓
$19.84 \mathrm{cal}$
C
$13.90 \mathrm{cal}$
D
$9.92 \mathrm{cal}$

Answer

Correct option: B.

$19.84 \mathrm{cal}$

$\Delta U=\mu C_V \Delta T$
$=2 \times 4.96 \times(342-340)$
$=19.84 \mathrm{cal}$

View full question & answer→

MCQ 1141 Mark

If $C_V=4.96 \mathrm{cal} /$ mole $K$, then increase in internal energy when temperature of $2$ moles of this gas is increased from $340 K$ to $342 K$

A
$27.80 \mathrm{cal}$
✓
$19.84 \mathrm{cal}$
C
$13.90 \mathrm{cal}$
D
$9.92 \mathrm{cal}$

Answer

Correct option: B.

$19.84 \mathrm{cal}$

$\Delta U=\mu C_V \Delta T$
$=2 \times 4.96 \times(342-340)$
$=19.84 \mathrm{cal}$

View full question & answer→

MCQ 1151 Mark

If $\Delta Q$ and $\Delta W$ represent the heat supplied to the system and the work done on the system respectively, then the first law of thermodynamics can be written as where $\Delta U$ is the internal energy

A
$\Delta Q=\Delta U+\Delta W$
✓
$\Delta Q=\Delta U-\Delta W$
C
$\Delta Q=\Delta W-\Delta U$
D
$\Delta Q=-\Delta W-\Delta U$

Answer

Correct option: B.

$\Delta Q=\Delta U-\Delta W$

From FLOT $\Delta Q=\Delta U+\Delta W$
$\because$ Heat supplied to the system so $\Delta Q \rightarrow$ Positive and work is done on the system so $\Delta W \rightarrow$ Negative.
Hence $+\Delta Q=\Delta U-\Delta W$

View full question & answer→

MCQ 1161 Mark

If $\Delta Q$ and $\Delta W$ represent the heat supplied to the system and the work done on the system respectively, then the first law of thermodynamics can be written as where $\Delta U$ is the internal energy

A
$\Delta Q=\Delta U+\Delta W$
✓
$\Delta Q=\Delta U-\Delta W$
C
$\Delta Q=\Delta W-\Delta U$
D
$\Delta Q=-\Delta W-\Delta U$

Answer

Correct option: B.

$\Delta Q=\Delta U-\Delta W$

From FLOT $\Delta Q=\Delta U+\Delta W$
$\because$ Heat supplied to the system so $\Delta Q \rightarrow$ Positive and work is done on the system so $\Delta W \rightarrow$ Negative.
Hence $+\Delta Q=\Delta U-\Delta W$

View full question & answer→

MCQ 1171 Mark

If $\Delta Q$ and $\Delta W$ represent the heat supplied to the system and the work done on the system respectively, then the first law of thermodynamics can be written as where $\Delta U$ is the internal energy

A
$\Delta Q=\Delta U+\Delta W$
✓
$\Delta Q=\Delta U-\Delta W$
C
$\Delta Q=\Delta W-\Delta U$
D
$\Delta Q=-\Delta W-\Delta U$

Answer

Correct option: B.

$\Delta Q=\Delta U-\Delta W$

From FLOT $\Delta Q=\Delta U+\Delta W$
$\because$ Heat supplied to the system so $\Delta Q \rightarrow$ Positive and work is done on the system so $\Delta W \rightarrow$ Negative.
Hence $+\Delta Q=\Delta U-\Delta W$

View full question & answer→

MCQ 1181 Mark

Unit mass of a liquid with volume $V_1$ is completely changed into a gas of volume $V_2$ at a constant external pressure $P$ and temperature $T$. If the latent heat of evaporation for the given mass is $L$, then the increase in the internal energy of the system is

A
Zero
B
$P\left(V_2-V_1\right)$
✓
$L-P\left(V_2-V_1\right)$
D
$L$

Answer

Correct option: C.

$L-P\left(V_2-V_1\right)$

$\Delta Q=\Delta V+P \Delta V $
$\Rightarrow m L=\Delta U+P(V-V)$
$\Rightarrow \Delta U=L-P(V-V) \quad(\because m=1)$

View full question & answer→

MCQ 1191 Mark

Unit mass of a liquid with volume $V_1$ is completely changed into a gas of volume $V_2$ at a constant external pressure $P$ and temperature $T$. If the latent heat of evaporation for the given mass is $L$, then the increase in the internal energy of the system is

A
Zero
B
$P\left(V_2-V_1\right)$
✓
$L-P\left(V_2-V_1\right)$
D
$L$

Answer

Correct option: C.

$L-P\left(V_2-V_1\right)$

$\Delta Q=\Delta V+P \Delta V $
$\Rightarrow m L=\Delta U+P(V-V)$
$\Rightarrow \Delta U=L-P(V-V) \quad(\because m=1)$

View full question & answer→

MCQ 1201 Mark

Unit mass of a liquid with volume $V_1$ is completely changed into a gas of volume $V_2$ at a constant external pressure $P$ and temperature $T$. If the latent heat of evaporation for the given mass is $L$, then the increase in the internal energy of the system is

A
Zero
B
$P\left(V_2-V_1\right)$
✓
$L-P\left(V_2-V_1\right)$
D
$L$

Answer

Correct option: C.

$L-P\left(V_2-V_1\right)$

$\Delta Q=\Delta V+P \Delta V $
$\Rightarrow m L=\Delta U+P(V-V)$
$\Rightarrow \Delta U=L-P(V-V) \quad(\because m=1)$

View full question & answer→

MCQ 1211 Mark

A thermodynamic process in which temperature $T$ of the system remains constant though other variable $P$ and $V$ may change, is called

A
lsochoric process
✓
Isothermal process
C
lsobaric process
D
None of these

Answer

Correct option: B.

Isothermal process

In isothermal process, temperature remains constant.

View full question & answer→

MCQ 1221 Mark

A thermodynamic process in which temperature $T$ of the system remains constant though other variable $P$ and $V$ may change, is called

A
lsochoric process
✓
Isothermal process
C
lsobaric process
D
None of these

Answer

Correct option: B.

Isothermal process

In isothermal process, temperature remains constant.

View full question & answer→

MCQ 1231 Mark

A thermodynamic process in which temperature $T$ of the system remains constant though other variable $P$ and $V$ may change, is called

A
lsochoric process
✓
Isothermal process
C
lsobaric process
D
None of these

Answer

Correct option: B.

Isothermal process

In isothermal process, temperature remains constant.

View full question & answer→

MCQ 1241 Mark

One mole of an ideal monoatomic gas is heated at a constant pressure of one atmosphere from $0^{\circ} \mathrm{C}$ to $100^{\circ} \mathrm{C}$. Then the change in the internal energy is

A
$6.56$ joules
B
$8.32 \times 10^2$ joules
✓
$12.48 \times 10^2$ joules
D
$20.80$ joules

Answer

Correct option: C.

$12.48 \times 10^2$ joules

Change in internal energy is always equal to the heat supplied at constant volume.
i.e. $\Delta U=(\Delta Q)_V=\mu C_V\Delta T$.
Form onoatomicgas $(C_V=\frac{3}{2R})$
$\Rightarrow \Delta U =\mu\left(\frac{3}{2} R\right) \Delta T$
$=1 \times\frac{3}{2}\times8.31\times(100-0)$
$ =12.48 \times 10^2 J$

View full question & answer→

MCQ 1251 Mark

One mole of an ideal monoatomic gas is heated at a constant pressure of one atmosphere from $0^{\circ} \mathrm{C}$ to $100^{\circ} \mathrm{C}$. Then the change in the internal energy is

A
$6.56$ joules
B
$8.32 \times 10^2$ joules
✓
$12.48 \times 10^2$ joules
D
$20.80$ joules

Answer

Correct option: C.

$12.48 \times 10^2$ joules

Change in internal energy is always equal to the heat supplied at constant volume.
i.e. $\Delta U=(\Delta Q)_V=\mu C_V\Delta T$.
Form onoatomicgas $(C_V=\frac{3}{2R})$
$\Rightarrow \Delta U =\mu\left(\frac{3}{2} R\right) \Delta T$
$=1 \times\frac{3}{2}\times8.31\times(100-0)$
$ =12.48 \times 10^2 J$

View full question & answer→

MCQ 1261 Mark

One mole of an ideal monoatomic gas is heated at a constant pressure of one atmosphere from $0^{\circ} \mathrm{C}$ to $100^{\circ} \mathrm{C}$. Then the change in the internal energy is

A
$6.56$ joules
B
$8.32 \times 10^2$ joules
✓
$12.48 \times 10^2$ joules
D
$20.80$ joules

Answer

Correct option: C.

$12.48 \times 10^2$ joules

Change in internal energy is always equal to the heat supplied at constant volume.
i.e. $\Delta U=(\Delta Q)_V=\mu C_V\Delta T$.
Form onoatomicgas $(C_V=\frac{3}{2R})$
$\Rightarrow \Delta U =\mu\left(\frac{3}{2} R\right) \Delta T$
$=1 \times\frac{3}{2}\times8.31\times(100-0)$
$ =12.48 \times 10^2 J$

View full question & answer→

MCQ 1271 Mark

A container that suits the occurrence of an isothermal process should be made of

✓
Copper
B
Glass
C
Wood
D
Cloth

Answer

Correct option: A.

Copper

An isothermal process takes place at constant temperature, must be carried out in a vessel with conducting wall so that heat generated should go out at once.

View full question & answer→

MCQ 1281 Mark

A container that suits the occurrence of an isothermal process should be made of

✓
Copper
B
Glass
C
Wood
D
Cloth

Answer

Correct option: A.

Copper

An isothermal process takes place at constant temperature, must be carried out in a vessel with conducting wall so that heat generated should go out at once.

View full question & answer→

MCQ 1291 Mark

A container that suits the occurrence of an isothermal process should be made of

✓
Copper
B
Glass
C
Wood
D
Cloth

Answer

Correct option: A.

Copper

An isothermal process takes place at constant temperature, must be carried out in a vessel with conducting wall so that heat generated should go out at once.

View full question & answer→

MCQ 1301 Mark

A carnot engine has the same efficiency between $800 K$ to $500 K$ and $x K$ to $600 K$. The value of $x$ is

A
$1000 \mathrm{~K}$
✓
$960 K$
C
$846 K$
D
$754 K$

Answer

Correct option: B.

$960 K$

In first case, $\left(\eta_1\right)=1-\frac{500}{800}=\frac{3}{8}$ and
in second case, $\left(\eta_2\right)=1-\frac{600}{x}$
Since $\eta_1=\eta_2$,
therefore $\frac{3}{8}=1-\frac{600}{x}$
$\text { or } \frac{600}{x}=1-\frac{3}{8}=\frac{5}{8} \text { or }x=\frac{600 \times 8}{5}=960 \mathrm{~K}$

View full question & answer→

MCQ 1311 Mark

A carnot engine has the same efficiency between $800 K$ to $500 K$ and $x K$ to $600 K$. The value of $x$ is

A
$1000 \mathrm{~K}$
✓
$960 K$
C
$846 K$
D
$754 K$

Answer

Correct option: B.

$960 K$

In first case, $\left(\eta_1\right)=1-\frac{500}{800}=\frac{3}{8}$ and
in second case, $\left(\eta_2\right)=1-\frac{600}{x}$
Since $\eta_1=\eta_2$,
therefore $\frac{3}{8}=1-\frac{600}{x}$
$\text { or } \frac{600}{x}=1-\frac{3}{8}=\frac{5}{8} \text { or }x=\frac{600 \times 8}{5}=960 \mathrm{~K}$

View full question & answer→

MCQ 1321 Mark

A carnot engine has the same efficiency between $800 K$ to $500 K$ and $x K$ to $600 K$. The value of $x$ is

A
$1000 \mathrm{~K}$
✓
$960 K$
C
$846 K$
D
$754 K$

Answer

Correct option: B.

$960 K$

In first case, $\left(\eta_1\right)=1-\frac{500}{800}=\frac{3}{8}$ and
in second case, $\left(\eta_2\right)=1-\frac{600}{x}$
Since $\eta_1=\eta_2$, therefore $\frac{3}{8}=1-\frac{600}{x}$
$\text { or } \frac{600}{x}=1-\frac{3}{8}=\frac{5}{8}$
$\text { or }x=\frac{600 \times 8}{5}=960 \mathrm{~K}$

View full question & answer→

MCQ 1331 Mark

A gas is suddenly compressed to one fourth of its original volume. What will be its final pressure, if its initial pressure is $P$

A
Lesss than $P$
✓
More than $P$
C
$P$
D
Either or

Answer

Correct option: B.

More than $P$

$P V^\gamma=\text { constant } $
$\Rightarrow \frac{P_2}{P_1}=\left(\frac{V_1}{V_2}\right)^\gamma=\left(\frac{V_1}{V_1 / 4}\right)^\gamma=4^\gamma $
$\Rightarrow P_2=4^\gamma P$
As $\gamma$ is always greater than one so $4^\gamma>4 \Rightarrow P_2>4 P$

View full question & answer→

MCQ 1341 Mark

A gas is suddenly compressed to one fourth of its original volume. What will be its final pressure, if its initial pressure is $P$

A
Lesss than $P$
✓
More than $P$
C
$P$
D
Either or

Answer

Correct option: B.

More than $P$

$P V^\gamma=\text { constant } $
$\Rightarrow \frac{P_2}{P_1}=\left(\frac{V_1}{V_2}\right)^\gamma=\left(\frac{V_1}{V_1 / 4}\right)^\gamma=4^\gamma $
$\Rightarrow P_2=4^\gamma P$
As $\gamma$ is always greater than one so $4^\gamma>4 \Rightarrow P_2>4 P$

View full question & answer→

MCQ 1351 Mark

A gas is suddenly compressed to one fourth of its original volume. What will be its final pressure, if its initial pressure is $P$

A
Lesss than $P$
✓
More than $P$
C
$P$
D
Either or

Answer

Correct option: B.

More than $P$

$P V^\gamma=\text { constant }$
$ \Rightarrow \frac{P_2}{P_1}=\left(\frac{V_1}{V_2}\right)^\gamma=\left(\frac{V_1}{V_1 / 4}\right)^\gamma=4^\gamma $
$\Rightarrow P_2=4^\gamma P$
As $\gamma$ is always greater than one so $4^\gamma>4 \Rightarrow P_2 > 4 P$

View full question & answer→

MCQ 1361 Mark

If $300 \mathrm{ml}$ of a gas at $27^{\circ} \mathrm{C}$ is cooled to $7^{\circ} \mathrm{C}$ at constant pressure, then its final volume will be

A
$540 \mathrm{ml}$
B
$350 \mathrm{ml}$
✓
$280 \mathrm{ml}$
D
$135 \mathrm{ml}$

Answer

Correct option: C.

$280 \mathrm{ml}$

$V \propto T$ at constant pressure
$\Rightarrow \frac{V_1}{V_2}=\frac{T_1}{T_2}$
$\Rightarrow V_2=\frac{V_1 T_2}{T_1}$
$=\frac{300 \times 280}{300}$
$=280 \mathrm{ml} \text {. }$

View full question & answer→

MCQ 1371 Mark

If $300 \mathrm{ml}$ of a gas at $27^{\circ} \mathrm{C}$ is cooled to $7^{\circ} \mathrm{C}$ at constant pressure, then its final volume will be

A
$540 \mathrm{ml}$
B
$350 \mathrm{ml}$
✓
$280 \mathrm{ml}$
D
$135 \mathrm{ml}$

Answer

Correct option: C.

$280 \mathrm{ml}$

$V \propto T$ at constant pressure $\Rightarrow \frac{V_1}{V_2}=\frac{T_1}{T_2}$
$\Rightarrow V_2=\frac{V_1 T_2}{T_1}$
$=\frac{300 \times 280}{300}$
$=280 \mathrm{ml} \text {. }$

View full question & answer→

MCQ 1381 Mark

If $300 \mathrm{ml}$ of a gas at $27^{\circ} \mathrm{C}$ is cooled to $7^{\circ} \mathrm{C}$ at constant pressure, then its final volume will be

A
$540 \mathrm{ml}$
B
$350 \mathrm{ml}$
✓
$280 \mathrm{ml}$
D
$135 \mathrm{ml}$

Answer

Correct option: C.

$280 \mathrm{ml}$

$V \propto T$ at constant pressure
$\Rightarrow \frac{V_1}{V_2}=\frac{T_1}{T_2}$
$\Rightarrow V_2=\frac{V_1 T_2}{T_1}$
$=\frac{300 \times 280}{300}$
$=280 \mathrm{ml} \text {. }$

View full question & answer→

MCQ 1391 Mark

The latent heat of vaporisation of water is $2240 \mathrm{~J} / \mathrm{gm}$. If the work done in the process of expansion of $1 \mathrm{~g}$ is $168 \mathrm{~J}$, then increase in internal energy is

A
$2408 \mathrm{~J}$
B
$2240 \mathrm{~J}$
✓
$2072 \mathrm{~J}$
D
$1904 \mathrm{~J}$

Answer

Correct option: C.

$2072 \mathrm{~J}$

$\Delta Q=\Delta U+\Delta W$
$ \Rightarrow \Delta U=\Delta Q-\Delta W$
$=2240-168$
$=2072\mathrm{~J}$.

View full question & answer→

MCQ 1401 Mark

The latent heat of vaporisation of water is $2240 \mathrm{~J} / \mathrm{gm}$. If the work done in the process of expansion of $1 \mathrm{~g}$ is $168 \mathrm{~J}$, then increase in internal energy is

A
$2408 \mathrm{~J}$
B
$2240 \mathrm{~J}$
✓
$2072 \mathrm{~J}$
D
$1904 \mathrm{~J}$

Answer

Correct option: C.

$2072 \mathrm{~J}$

$\Delta Q=\Delta U+\Delta W$
$ \Rightarrow \Delta U=\Delta Q-\Delta W$
$=2240-168$
$=2072\mathrm{~J}$.

View full question & answer→

MCQ 1411 Mark

The latent heat of vaporisation of water is $2240 \mathrm{~J} / \mathrm{gm}$. If the work done in the process of expansion of $1 \mathrm{~g}$ is $168 \mathrm{~J}$, then increase in internal energy is

A
$2408 \mathrm{~J}$
B
$2240 \mathrm{~J}$
✓
$2072 \mathrm{~J}$
D
$1904 \mathrm{~J}$

Answer

Correct option: C.

$2072 \mathrm{~J}$

$\Delta Q=\Delta U+\Delta W $
$\Rightarrow \Delta U=\Delta Q-\Delta W$
$=2240-168$
$=2072\mathrm{~J}$.

View full question & answer→

MCQ 1421 Mark

A measure of the degree of disorder of a system is known as

A
Isobaric
B
Isotropy
C
Enthalpy
✓
Entropy

Answer

Correct option: D.

Entropy

View full question & answer→

MCQ 1431 Mark

A measure of the degree of disorder of a system is known as

A
Isobaric
B
Isotropy
C
Enthalpy
✓
Entropy

Answer

Correct option: D.

Entropy

View full question & answer→

MCQ 1441 Mark

A measure of the degree of disorder of a system is known as

A
Isobaric
B
Isotropy
C
Enthalpy
✓
Entropy

Answer

Correct option: D.

Entropy

View full question & answer→

MCQ 1451 Mark

The process in which no heat enters or leaves the system is termed as

A
Isochoric
B
Isobaric
C
lsothermal
✓
Adiabatic

Answer

Correct option: D.

Adiabatic

In adiabatic process, no transfer of heat takes place between system and surrounding.

View full question & answer→

MCQ 1461 Mark

The process in which no heat enters or leaves the system is termed as

A
Isochoric
B
Isobaric
C
lsothermal
✓
Adiabatic

Answer

Correct option: D.

Adiabatic

In adiabatic process, no transfer of heat takes place between system and surrounding.

View full question & answer→

MCQ 1471 Mark

The process in which no heat enters or leaves the system is termed as

A
Isochoric
B
Isobaric
C
lsothermal
✓
Adiabatic

Answer

Correct option: D.

Adiabatic

In adiabatic process, no transfer of heat takes place between system and surrounding.

View full question & answer→

MCQ 1481 Mark

During an adiabatic expansion of $2$ moles of a gas, the change in internal energy was found $-50.$ The work done during the process is

A
Zero
B
$100$ J
C
$-50$ J
✓
$50$ J

Answer

Correct option: D.

$50$ J

For adiabatic forces $\Delta W=-\Delta U \quad(\because \Delta Q=0)$
$\Rightarrow \Delta W=-(-50)=+50 \mathrm{~J}$

View full question & answer→

MCQ 1491 Mark

During an adiabatic expansion of $2$ moles of a gas, the change in internal energy was found $-50.$ The work done during the process is

A
Zero
B
$100$ J
C
$-50$ J
✓
$50$ J

Answer

Correct option: D.

$50$ J

For adiabatic forces $\Delta W=-\Delta U \quad(\because \Delta Q=0)$
$\Rightarrow \Delta W=-(-50)=+50 \mathrm{~J}$

View full question & answer→

MCQ 1501 Mark

During an adiabatic expansion of $2$ moles of a gas, the change in internal energy was found $-50.$ The work done during the process is

A
Zero
B
$100$ J
C
$-50$ J
✓
$50$ J

Answer

Correct option: D.

$50$ J

For adiabatic forces $\Delta W=-\Delta U \quad(\because \Delta Q=0)$
$\Rightarrow \Delta W=-(-50)=+50 \mathrm{~J}$

View full question & answer→

JEE physics MCQ