Ask & Answer Question

Physics NCERT Exemplar Solutions Class 11th Chapter Five Physics Class 12th Physics NCERT Exemplar Solutions Class 12th Chapter Twelve Thermodynamics

A cylinder of fixed capacity of 44.8 litres contains helium gas at standard gas at standard temperature and pressure. The amount of heat needed to raise the temperature of gas in the cylinder by 20.0°C will be:

(Given gas constant R = 8.3 JK^-¹ -mol^-¹)

Option 1 -

249 J

Option 2 -

415 J

Option 3 -

498 J

Option 4 -

830 J

2 Views | Posted 2 months ago

Asked by Shiksha User

1 Answer

Answered by

alok kumar singh | Contributor-Level 10

2 months ago

Correct Option - 3

Detailed Solution:

44.4 litres of He $\frac{44.8}{22.4} m o l e s$

$η = 2$ moles of He

Q = $η C_{v} Δ T$ (for fixed capacity (v = constant)

= $2 \times \frac{R}{γ - 1} Δ T$

= $\frac{2 \times R}{\frac{5}{3} - 1} \times 20$

= $= 2 \times \frac{3 R}{2} \times 20$

Q = 60R = 60 × 8.3

Q = 498 J

44.4 litres of He <math> <mrow> <mfrac> <mrow> <mn>4</mn> <mn>4</mn> <mo>.</mo> <mn>8</mn> </mrow> <mrow> <mn>2</mn> <mn>2</mn> <mo>.</mo> <mn>4</mn> </mrow> </mfrac> <mi>m</mi> <mi>o</mi> <mi>l</mi> <mi>e</mi> <mi>s</mi> </mrow> </math>             <math> <mrow> <mi>η</mi> <mo>=</mo> <mn>2</mn> </mrow> </math>   <!- [if gte mso 9]><xml> <o:OLEObject Type="Embed" ProgID="Equation.DSMT4" ShapeID="_x0000_i1026" DrawAspect="Content" ObjectID="_1814258358"> </o:OLEObject></xml><! [endif]-> moles of He              Q = <!- [if gte mso 9]><xml> <o:OLEObject Type="Embed" ProgID="Equation.DSMT4" ShapeID="_x0000_i1027" DrawAspect="Content" ObjectID="_1814258359"> </o:OLEObject></xml><! [endif]->  <math> <mrow> <mi>η</mi> <msub> <mrow> <mi>C</mi> </mrow> <mrow> <mi>v</mi> </mrow> </msub> <mi>Δ</mi> <mi>T</mi> </mrow> </math> (for fixed capacity (v = constant)              = <!- [if gte mso 9]><xml> <o:OLEObject Type="Embed" ProgID="Equation.DSMT4" ShapeID="_x0000_i1028" DrawAspect="Content" ObjectID="_1814258360"> </o:OLEObject></xml><! [endif]->  <math> <mrow> <mn>2</mn> <mo>×</mo> <mfrac> <mrow> <mi>R</mi> </mrow> <mrow> <mi>γ</mi> <mo>−</mo> <mn>1</mn> </mrow> </mfrac> <mi>Δ</mi> <mi>T</mi> </mrow> </math>               = <!- [if gte mso 9]><xml> <o:OLEObject Type="Embed" ProgID="Equation.DSMT4" ShapeID="_x0000_i1029" DrawAspect="Content" ObjectID="_1814258361"> </o:OLEObject></xml><! [endif]->  <math> <mrow> <mfrac> <mrow> <mn>2</mn> <mo>×</mo> <mi>R</mi> </mrow> <mrow> <mfrac> <mrow> <mn>5</mn> </mrow> <mrow> <mn>3</mn> </mrow> </mfrac> <mo>−</mo> <mn>1</mn> </mrow> </mfrac> <mo>×</mo> <mn>2</mn> <mn>0</mn> </mrow> </math> = <math> <mrow> <mo>=</mo> <mn>2</mn> <mo>×</mo> <mfrac> <mrow> <mn>3</mn> <mi>R</mi> </mrow> <mrow> <mn>2</mn> </mrow> </mfrac> <mo>×</mo> <mn>2</mn> <mn>0</mn> </mrow> </math>  Q = 60R = 60 × 8.3  Q = 498 J

0 Upvotes 0 Downvotes

Similar Questions for you

Starting with the same initial conditions, an ideal gas expands form volume V₁ to V₂ in the three different ways. The work done by the gas in W₁ if the process is purely isothermal, W₂, if the process is purely adiabatic and W₃ if the process is purely isobaric. Then, choose the correct option

Vishal Baghel

Process-AB $\Rightarrow$ Isobaric,

Process-AC $\Rightarrow$ Isothermal, and

Process-ADÞAdiabatic

W₂ < W₁ < W₃

In 5 minutes, a body cools from 75°C to 65°C at room temperature of 25°C. The temperature of body at the end of next 5 minutes is__________°C.

Vishal Baghel

from newtan’s law of cooling :

$\frac{d T}{d t} = - k (T - T_{0})$

Using average value :

$\frac{75 - 65}{5} = - k [\frac{(75 + 65) 2}{2} - 25]$

$\frac{(65 - T)}{5} = - k [\frac{(65 + T)}{2} - 25]$

$\Rightarrow \frac{10}{65 - T} = \frac{45}{\frac{65 + T}{2} - 25}$

$\Rightarrow T = 57 ° C$

A Carnot engine has efficiency of 50%. If the temperature of sink is reduced by 40°C, its efficiency increases by 30%. The temperature of the source will be:

Vishal Baghel

Efficiency $η = 50 %$

$\Rightarrow η = (1 - \frac{T_{2}}{T_{1}}) \times 100 = 50$

$\Rightarrow 1 - \frac{T_{2}}{T_{1}} = \frac{50}{100} \Rightarrow 1 - \frac{T_{2}}{T_{1}} = \frac{1}{2}$ -(i)

After the change :- $1 - \frac{T_{2}^{1}}{T_{1}} = \frac{50 \times (1.3)}{100} = \frac{65}{100} = . 65$ $\frac{_{}^{}}{_{}} \frac{}{} \frac{}{}$ -(ii)

$\Rightarrow 1 - (\frac{T_{2} - 40}{T_{1}}) = . 65$ --(ii)

$\Rightarrow \frac{40}{T_{1}} = . 65 - . 50 = . 15 \Rightarrow T_{1} = \frac{4000}{15} = 266.7 k$

A 110 V, 50Hz, AC source is connected in the circuit (as shown in figure). The current through the resistance $55 Ω,$ at resonance in the circuit, will be__________A.

Payal Gupta

At Resonance

X_L = X_C

then l_L = l_C

Now phasor diagram

for L & C

So, Net current = zero

$= (l_{L} + l_{C})$

Therefore current through R circuit at resonance will be zero

A proton and an alpha particle of the same velocity enter in a uniform magnetic field which is acting perpendicular to their direction of motion. The ratio of the radii of the circular paths described by the alpha particle and proton is

Payal Gupta

$r = \frac{m v}{q B}$

$r_{α} = \frac{m_{α} v}{q_{α} B}$

$r_{P} = \frac{m_{P} v}{q_{P} B}$

$\frac{r_{α}}{r_{P}} = \frac{m_{α} q_{P}}{m_{P} q_{α}} = \frac{m_{α}}{m_{P}} (\frac{q_{P}}{q_{α}})$

$= \frac{4 m}{m} (\frac{q}{2 q}) = 2 : 1$

Get authentic answers from experts, students and alumni that you won't find anywhere else

On Shiksha, get access to

65k Colleges
1.2k Exams
688k Reviews
1800k Answers

Community GuidelinesUser Points System

QuestionsDiscussionsTags

A cylinder of fixed capacity of 44.8 litres contains helium gas at standard gas at standard temperature and pressure. The amount of heat needed to raise the temperature of gas in the cylinder by 20.0°C will be:

(Given gas constant R = 8.3 JK^-¹ -mol^-¹)

1 Answer

Similar Questions for you

Learn more about...

Most viewed information

Most viewed information

Share Your College Life Experience

Didn't find the answer you were looking for?

Need guidance on career and education? Ask our experts

Your Question

A cylinder of fixed capacity of 44.8 litres contains helium gas at standard gas at standard temperature and pressure. The amount of heat needed to raise the temperature of gas in the cylinder by 20.0°C will be: (Given gas constant R = 8.3 JK-1 -mol-1)

1 Answer

Similar Questions for you

Learn more about...

Most viewed information

Most viewed information

Share Your College Life Experience

Didn't find the answer you were looking for?

Need guidance on career and education? Ask our experts

Your Question

A cylinder of fixed capacity of 44.8 litres contains helium gas at standard gas at standard temperature and pressure. The amount of heat needed to raise the temperature of gas in the cylinder by 20.0°C will be:

(Given gas constant R = 8.3 JK^-¹ -mol^-¹)