Question

The work done in heating one mole of an ideal gas at constant pressure from 15 degree Celsius to 25 degree Celsius?
1) 1.987 cal
2) 198.7 cal
3) 9.935 cal
4) 19.87 cal

Answer 1

Answer:

No. of moles of the gas, n = 1 mole

Temperature,

T₁ = 15 deg celcius = 15 + 273 = 288 K

and

T₂ = 25 deg celcius = 25 + 273 = 298 K

The Ideal gas constant, R = 8.314 J K⁻¹ mol⁻¹

Thus,  

The work done in heating 1 mole of gas is,

= n R ∆ T

= n R (T₂ – T₁)

= 1 * 8.314 * (298 - 288)

= 8.314 * 10

= 83.14 Joules

= 83.14 / 4.184 ….. [∵ 1 joule = 4.184 cal]

= 19.87 cal

Answer 2

Answer:

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